Chemistry, 26.02.2020 19:52 carterh166

At 473 K, for the elementary reaction 2NOCl(g)k1⇌k−12NO(g)+Cl2(g) k1=7.8×10−2L/mols and k−1=4.7×102L2/mol2s A sample of NACl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.58 mol/L .

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Answer from: naomicervero

Part a)

[NO] = 0.048 mol/L

Part b)

[Cl₂] = 0.024 mol/L

Explanation:

The complete question is:

At 473 K, for the elementary reaction 2NOCl(g) ⇌2NO(g)+Cl₂(g)

k₁ = 7.8×10⁻² L/mol.s and k⁻¹ =4.7×10²L²/mol².s .

A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.58 mol/L.

Part a) Calculate the concentration of NO

Part b) Calculate the concentration of Cl₂

Solution

1. Calculate the equilibrium constant.

k₁ is the rate constant of the forward reaction and k⁻¹ is the rate constant of the reverse reaction.

The equilibrium constant is the quotient of the rate constant for the forward reaction and the rate constant of the reverse reaction:

$K_{eq}=\dfrac{K}{k^{-1}}=\dfrac{7.8\times 10^{-2}L/mol.s}{4.7\times 10^2L^2/mol^2.s}\\\\\\ K_{eq}=1.6596\times 10^{-4}mol/L$

2. Build the ICE (Initial, Change, Equilibrium) table

2NOCl(g) ⇌ 2NO(g) + Cl₂(g)

Initial A 0 0

Change -2x +2x +x

Equilibrium A - 2x 2x x

A - 2x = 0.58 mol/L

3. Write the equilibrium constant equation

$K_{eq}=\dfrac{{[NO]}^2].[Cl_2]}{[NOCl]^2}\\ \\ \\ 1.6593\times 10^{-4}mol/L=\dfrac{(2x)^2(x)}{(0.58mol/L)^2}$

4. Solve the equation

$4x^3=5.5828\times 10^{-5}\\\\\\x=0.0241mol/L$

5. Concentrations

Part a)

[NO] = 2x = 2(0.0241mol/L) = 0.048 mol/L

Part b)

[Cl₂] = x = 0.024 mol/L

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At 473 K, for the elementary reaction 2NOCl(g)k1⇌k−12NO(g)+Cl2(g) k1=7.8×10−2L/mols and k−1=4.7×102L...

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