At 35°C, K = 1.6105 for the following reaction. 2 NOCl(g) equilibrium reaction a

Question

At 35°C, K = 1.6105 for the following reaction. 2 NOCl(g) equilibrium reaction arrow 2 NO(g) + Cl2(g) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. 2 NOCl(g) equilibrium reaction arrow 2 NO(g) + Cl2(g)

(a) 2.5 mol of pure NOCl in a 2.2-L flask NOCL NO Cl2

(b) 7.8 mol of NO and 3.9 mol of Cl2 in a 1.0-L flask NOCL NO Cl2

(c) 1.7 mol of NOCl and 1.7 mol of NO in a 1.8-L flask NOCl NO Cl2

(d) 3.9 mol of NO and 1.3 mol of Cl2 in a 1.0-L flask NOCl NO Cl2

(e) 3.0 mol of NOCl, 3.0 mol of NO, and 1.5 mol of Cl2 in a 1.5-L flask NOCl NO Cl2

(f) 1.13 mol/L concentration of all three gases NOCl NO Cl2

Explanation / Answer

2NOCl(g) <---> 2NO(g) + Cl2(g)

K = [NO]^2[Cl2]/[NOCl]^2

1.6*10^(-5) = [NO]^2[Cl2]/[NOCl]^2

a) Concentration of NOCl = 2.5/2.2 = 1.14 M

Concentration of NO = 2x = 0.006 M

Concentration of Cl2 = x = 0.003 M

1.6*10^(-5) = 2x*x/(1.14-x)^2

x = 0.003 M

b)

Concentration of NOCl = 2x =

Concentration of NO = 7.8/1= 7.8-2x =   M

Concentration of Cl2 = 3.9/1 = 3.9-x = M

1.6*10^(-5) = ((7.8-2x)^2(3.9-x))/(2x)^2

solve for x you will get all

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