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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=4.7×102L2/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 .

Answer :

Edufirst

Answer:

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:

               [tex]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[/tex]

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

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

4. Solve the equation

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

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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