Consider the following gas phase reaction at 25°C. N2(g) + C2H2(g) 2HCN(g) 1.600

Question

Consider the following gas phase reaction at 25°C.

N2(g) + C2H2(g) 2HCN(g)

1.600 mol N2(g) and 1.750 mol C2H2(g) are placed in a 1.000 L vessel and the mixture is allowed to react. At equilibrium, there are 1.587 mol N2(g) in the mixture.

a) What is Kc for this reaction at 25 °C?

b) If some amount of C2H2 is added to the system at equilibrium, how does the concentration of HCN will change when equilibrium is reinstated?

If someone could show me how to do this using the ICE table that would be great :) Thanks

Explanation / Answer

N2(g) + C2H2(g) 2HCN(g)

n = 1.6 mol of N2

n = 1.75 mol of C2H2

n = 0 of HCN

assume n = M since V = 1 L

M = 1.587 mol of N2 in equilibrium

ICE TABLE

Substance N2 C2H2 HCN

Initial Cnocnetration 1.6 1.75 0

Change (sotihiometric) -x -x +2x

Equilibrium Concnetration 1.6-x 1.75-x 2x

Which are here:

Write the concnetrations in equilibrium

[N2] = 1.6 - x

[C2H2] = 1.75 - x = 1.587

[HCN] = 0 + 2x

Solve for x (from C2H2)

1.75 - x = 1.587

x = 0.163

Substitute in all other concentrations

[N2] = 1.6 - x = 1.6 - 0.163 = 1.437

[HCN] = 0 + 2x = 2*0.163 = 0.326

Now, write the K expression

K = [HCN]^2 / [N2][C2H2]

K = (0.326)^2 / (1.587 * 1.437) = 0.01519

b)

if you add C2H2 ...

The concnetration of HCN will increase

since the shift goes from reactants to products

N2 + C2H2 ----> 2HCN

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