The equilibrium constant for A + 2B → 3C is 2.1 * 10^-6
Determine the equilibrium constant for 2A + 4B → 6C.
a- 4.2 * 10^-6
b- 4.4 * 10^-12
c- 2.3 *10^11
d- 1.8 *10^-11
e- None of these

Answers

Answer 1

Answer:

b- 4.4 * 10^-12.

Explanation:

Hello.

In this case, as the reaction:

A + 2B → 3C

Has an equilibrium expression of:

[tex]K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}[/tex]

If we analyze the reaction:

2A + 4B → 6C

Which is twice the initial one, the equilibrium expression is:

[tex]K_2=\frac{[C]^6}{[A]^2[B]^4}[/tex]

It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:

[tex]K_2=K_1^2[/tex]

Thus, the equilibrium constant of the second reaction turns out:

[tex]K_2=(2.1 * 10^{-6})^2\\\\K_2=4.4x10^{-12}[/tex]

Therefore, the answer is b- 4.4 * 10^-12.

Best regards.


Related Questions

An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 74.6 kJ of heat. Before the reaction, the volume of the system was 8.20 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.

Answers

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

True or False: The exact location of an electron can be measured thanks to
modern science.

Answers

Answer:

false you can not get a exact location of electrons from just modern science

How many atoms are in 10 g of He

Answers

Answer:

6.7

10

23

atoms of H

Explanation:

convert 575.1 mmHg to atm

Answers

Answer:

= .7567105263

Explanation:

1 atm = 760 mmHg

575.1 mmHg (1 atm/760mmHg) = .7567105263 atm

A scientist discovers remnants of an organism on a slide under his microscope. He can only identify a few components: a large vacuole, ridged cell wall, and a chloroplast. Was the organism prokaryotic or eukaryotic? How do you know?

Answers

Answer:

Eukaryotic (a plant cell)

Explanation:

The presence of a chloroplast indicates that the cell has membrane-bound organelles. This is not a feature of prokaryotic cells - only eukaryotic cells. Therefore, the cell is eukaryotic. The presence of a large vacuole, chloroplast, and rigid cell wall suggests its a plant cell as plant cells are the only eukaryotes with these features.

When a helium balloon rises in the air, it expands. If the volume of the balloon doubles, what happens to the density of the helium inside it?
a.The density decreases by half
b.The density doubles
c.The density triples
d.The density stays the same

Answers

A. The density decreases by half
A. the density decreases by half

Calculate the theoretical density (in g/cm3) of copper (Cu), given that it has the FCC structure. The atomic weight of Cu is 63.55 g/mol, and its atomic radius R is 0.1278 nm.

Answers

Answer:

8.937g/cm³

Explanation:

To answer this question we need to know that, in 1 unit FCC cell you have:

Edge length = √8 * R

Volume = 8√8 * R³

And there are 4 atoms per unit cell

Mass of 4 atoms in g:

4 atom * (1mol / 6.022x10²³atom) * (63.55g / mol) = 4.221x10⁻²²g

Volume in cm³:

0.1278nm * (1x10⁻⁷cm / 1nm) = 1.278x10⁻⁸cm

Volume = 8√8 * (1.278x10⁻⁸cm)³

Volume = 4.723x10⁻²³cm³

And density is:

4.221x10⁻²²g / 4.723x10⁻²³cm³ =

8.937g/cm³






Which number represetns a coefficient?
2
3
4
7

Answers

4 is the number that represents a coefficient

Solid calcium chlorate decomposes to form solid calcium chloride and oxygen gas.

Write the balanced chemical equation for the reaction described. Phases are optional.

equation:

Answers

Answer:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Explanation:

Chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Balance chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Step 1:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 2

Step 2:

 Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 6

HELPP

describe what potassium would do to be more stable

Answers

Answer:

Explanation:

Its a elemental potassium is soft ,white  in colour  and has one more electron than argon,an element that we know is extremely stable ... Potassium extra electron is easily lost to form the much more stable cation, K+

the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?

Answers

Answer:

768g

Explanation:

We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.

The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram  were in the original sample.

What is half life?

The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.

The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.

ln P = kt + C

P = amount of I-137 at time t

C = constant

k = 1/time

t = time

1st condition:

P = Po, t = 0 days

2nd condition: (half-life)

P = 0.5Po, t = 8.07 days

3rd condition:

P = 25 grams, t = 40.35 days

Po = 800 grams

mass of I-137 = 800 gram

Therefore, 800 gram were in the original sample.

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Please help!!
This is a big part of my grade -----
Will make you brainliest******

Answers

Explanation:

U need to draw the graph first and make a line at 17 pennies, where the line of 17 pennies and your graph meet is the mass of it(at y axis)

How many moles are in 141.16 grams of F?
Use two digits past the decimal for all values

Answers

Answer: 2681.81

Explanation:

2681.81

hope that helps

Calculate the pH of a solution containing a caffeine concentration of 455 mg/L . Express your answer to one decimal place.

Answers

Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

               

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

The normal boiling point of benzene is 80.1°C. What is its enthalpy of vaporization if the vapor pressure at 26.1°C is 100 torr?

Answers

The answer is, 33.0 kJ/mol

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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What is the process of cell eating called

Answers

Answer:

Phagocytosis

Explanation:

Which of the following evidence supports the theory of plate tectonics

Answers

The theory of plate tectonics states that the Earth's solid outer crust, the lithosphere, is separated into plates that move over the asthenosphere, the molten upper portion of the mantle.,Thus, at divergent boundaries, oceanic crust is created that’s what plate tectonics means hoped that helped

Answer:

seafloor spreading

Explanation:

i took the test

Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.

Answers

Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6

Calculate the volume of the gas, in liters, if 1.75 mol has a pressure of 1.28 atm at a temperature of -7 ∘C

Answers

Answer:

A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.

 

You need to apply the ideal gas law PV=nRT

 

You have the pressure, P=1.01 atm

you have the volume, V = 2.21 L

The ideal gas constant R= 0.08205 L. atm/ mole.K at  273 K

 

find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)

 

n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and

L=1.23 L

 

P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L

= 1.994 atm

Explanation:

How is matter divided?

Answers

matter can be divided into 2 ways:

if it’s a pure substance it can be broken down into elements and compounds

and if it’s a mixture it could be physically combined with other structures to separate their original components

Hopefully i could help!!

The empirical formula of CBr2 has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound

Answers

Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

is carried out in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm3. If the equilibrium conversion is found to be 60%, (a) What is the equilibrium constant, KC if the reaction is a gas phase reaction? (Ans.: Gas: KC = 0.328 dm3/mol)

Answers

Answer:

0.328 mol/dm³

Explanation:

We have

I started this calculation from Rate's law.

Remember equilibrium constant has been given to be 60%

Our interest is Kc, that is the equilibrium constant.

Ca = 4(1-0.6)/1+(-0.5*0.6)

= 4-2.4/1-0.3

= 1.6/0.7

= 2.2857

Cb = 4x0.6/2(1+(-0.5*0.6))

= 2.4/2(0.7)

= 2.4/1.4

= 1.7143

Kc = Cb/Ca²

= 1.7143/2.2857²

= 1.7143/5.2244

= 0.328 mol/dm³

I have added an attachment showing earlier stages to the final answer

The following are placed in a beaker weighing 39.457 g:
2.689 g of NaCl, 1.26 g of sand and 5.0 g water
What is the final mass of the beaker?

Answers

Answer:

48.4 g

Explanation:

A chemist prepares a solution of silver(II) oxide by measuring out 0.0013 of silver(II) oxide into a 100 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's silver(II) oxide solution. Be sure your answer has the correct number of significant digits.

Answers

Answer:

1.3x10⁻⁸ mol/L

Explanation:

0.0013μmol, Calculate concentration in mol/L

To obtain concentration in mol/L we need to convert the μmoles to moles and mL to liters:

Moles silver(II) oxide:

0.0013μmol × (1mol / 1x10⁶μmol) = 1.3x10⁻⁹ moles

Liters solution:

100mL * (1L / 1000mL) = 0.1L

That means concentration in mol/L is:

1.3x10⁻⁹ moles / 0.1L =

1.3x10⁻⁸ mol/L

What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)

Answers

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Best regards!

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

Answers

Complete Question

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

(a) before addition of any HCl (b) after addition of 25.0 mL of HCl

Answer:

a The value  is  [tex]pH =12.81[/tex]

b [tex]pH  = 11.9[/tex]

Explanation:

From the question we are told that

  The first pKb value  for B is [tex]pK_b_1  =  2.10[/tex]

   The second pKb value  for B is [tex]pK_b_2  =  7.54[/tex]

     The volume is  [tex]V =   50.0 mL   =[/tex]

     The  concentration  of  B is  [tex][B]  =  0.60 M[/tex]

     The concentration of [tex]C_A =  0.60 M[/tex]

Generally the reaction equation showing the first dissociation of B is  

[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)}  +  OH^- _{(aq)} }[/tex]

Here the ionic  constant for B is mathematically represented as

      [tex]K_i  =  \frac{[BH^+] [OH^-]}{[B]}[/tex]

Let denot the concentration of  [BH^+]  as  z  and  since [tex][BH^+] =  [OH^-][/tex] then [tex][OH^-][/tex] is also  z

So  [B] =  0.60  -  z  

Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is

   [tex]K_i  =  7.94 *10^{-3}[/tex]

So

      [tex] 7.94 *10^{-3}=  \frac{z^2}{ 0.60 - z}[/tex]

=>   [tex]z^ 2 + 0.00794 z - 0.00476[/tex]

using quadratic formula to solve this equation

     [tex]z = 0.0651[/tex]

Hence the concentration of  [tex]OH^{-}[/tex] is   [tex][OH^-] =0.0651[/tex]

Generally  [tex]pOH =  -log [OH^-][/tex]

=>    [tex]pOH =  -log (0.065)[/tex]

=>    [tex]pOH = 1.187 [/tex]

Generally the pH is mathematically represented as

    [tex]pH = 14 - 1.187[/tex]

      [tex]pH =12.81[/tex]

Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is   [tex] 50 mL [/tex]

The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL

Now the pOH at half way to the first dissociation of the base is  

     [tex]pOH  =  -log(K_i)[/tex]

=>   [tex]pOH  =  -log(0.00794)[/tex]

=>   [tex]pOH  =  2.100[/tex]

Generally the pH after addition of 25.0 mL of HCl is  

    [tex]pH  =  14 -  2.100[/tex]\

=>   [tex]pH  = 11.9[/tex]

The first dissociation's equation is as follows:

[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]

The second dissociation of the base equation is

[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]

[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]

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The amount of force that is exerted on a balloon by the gas inside the balloon is.
O A) temperature
OB) prlessure
O C) volume
O D) heat

Answers

Answer:

pressure

Explanation:

pressure is the amount of force exerted on an area. when you blow up the balloon you're filling it with gas particles. the gas particles move freely within the balloon and may collide with one another exerting pressure on the inside of the balloon.

The pressure of the gas is the amount of force that is exerted on a balloon by the gas inside the balloon. Therefore, option B is correct.

What is pressure?

Pressure can be described as the force applied perpendicular to the surface of a body per unit area. Pressure can be defined as a standard mechanical quantity and is derived from a unit of force divided by a unit of area.

The SI unit of measurement of pressure, the pascal (Pa) or Newton per square meter (N/m²). Pressure can be defined as the amount of force exerted perpendicular to the surface per unit area.

Mathematically, the pressure exerted by force can be calculated as:

[tex]{\displaystyle p={\frac {F}{A}}}[/tex]

where, p is the pressure, F is the magnitude of the normal force, and A is the area of the surface.

Therefore, the amount of force that is exerted on the balloon by the gas inside the balloon is equal to pressure.

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Which of the following is NOT a strong electrolyte?
:

Answers:
Na2SO4
KI
CaCl2
LiOH
C6H1206

Answers

Answer:

C6H1206

Explanation:

C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.

What is the percent error for the experiment if the actual density is
2.49g/mL but the experimental value is 1.47 g/mL?

Answers

Answer:

The answer is 40.96%

Explanation:

The percentage error of a certain measurement can be found by using the formula

[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]

From the question

actual density = 2.49g/mL

error = 2.49 - 1.47 = 1.02

We have

[tex]p(\%) = \frac{1.02}{2.49} \times 100 \\ = 40.96385542...[/tex]

We have the final answer as

40.96 %

Hope this helps you

How does temperature affect the copper (II) chloride equilibrium? Is the forward reaction (color changing from blue to green) endothermic or exothermic? Justify your choice with experimental evidence i.e color changes in the video for Part B.

Answers

Answer:

See explanation

Explanation:

A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.

As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)

The equilibrum is represented by the equation;

[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<------>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive

The equilibrium mixture undergoing cooling or heating have colour changes. The temperature affects the colour of the products formed and the forward reaction is endothermic.

What are the equilibrium and the forward reactions?

In the reaction copper (II) chloride or [tex]\rm CuCl_{4}[/tex] is the main species. The heat or the temperature affects the colour formation of copper (II) chloride as the equilibrium change affects the colouration of the product.

The heating of the solution affects the colour change from blue to the green of the reactant to products and the forward reaction shifts the equilibrium towards the right when the temperature is increased and is an endothermic reaction.

The reaction at the equilibrium can be shown as,

[tex]\rm [Cu(H_{2}O)_{6}]^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow [CuCl_{4}]^{2-}(aq) + 6H_{2}O(l), \Delta H=positive[/tex]

Therefore, temperature changes the colouration and the forward reaction is an endothermic reaction.

Learn more about temperature and equilibrium here:

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