Answer to Question #207727 in Physical Chemistry for Aditya venkatesh

Question

1)Consider the following reaction: N2O4 (g) → 2 NO2 (g) Assume that an experiment is carried out in which the starting concentration of N2O4 (g) is 0.100 mol/L. No products are present at the beginning of the reaction. When equilibrium is established, the concentration of NO2 (g) is 2.43 x 10-2 mol/L. Calculate the equilibrium constant for the above reaction.

2)Nitrogen and oxygen react to produce nitric oxide according to the following equation: N2 (g) + O2 (g) → 2 NO (g) The equilibrium constant for this reaction is 1.70 x 10-3. Suppose that 0.920 mol N2 and 0.700 mol O2 are mixed in a 3.00-L reaction vessel. What will be the concentrations of N2, O2, and NO when equilibrium is established? (Hint: assume that the amounts of N2 and O2 that react are small—less than 10% of the starting amounts—and check your assumption when you have solved the equation.)

Answer 1

1)

$K=\frac{[NO_2]^2}{[N_2O_4]}=\frac{(2.43\times10^{-2})^2}{0.1-\frac{1}{2}2.43\times10^{-2}}=6.72\times{10^{-3}}$ mol/L

2)

$K=\frac{[NO]^2}{[N_2][O_2]}=1.70\times10^{-3} \\ [N_2]_0 = 0.920/3 = 0.3067 \ mol/L \\ [O_2]_0 = 0.2333 \ mol/L \\ Let \ the \ concentration \ of \ NO \ be \ x, then \\ 1.70\times10^{-3} = \frac{x^2}{(0.3067-0.5x)(0.2333-0.5x)} \\ x^2=1.70\times10^{-3}(0.3067-0.5x)(0.2333-0.5x) \\ solving \ the \ quadratic\ equotion \ for \ x \\ x = 0.0108 - concentration \ of \ NO\\ [N_2]=0.3067-0.0108/2 = 0.3013 \ mol/L\\ [O_2]=0.2333-0.0108/2=0.2279 \ mol/L$