Question

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g)Kc = 4.3 × 10−25 2NO(g) + O2(g) ⇌ 2NO2(g)Kc = 6.4 × 109 Determine the value of the equilibrium constant for the following equation at the same temperature: N2(g) + 2O2(g) ⇌ 2NO2(g) Kc = × 10 (Enter your answer in scientific notation.)

Answer 1

Answer:

$K_c=2.752\times 10^{-15}$

Explanation:

The given equilibrium reaction is:

$N_2_{(g)}+O_2_{(g)}\rightleftharpoons 2NO_{(g)}$

The expression for the equilibrium constant is:

$K_c_1=\frac {[NO]^2}{[N_2][O_2]}=4.3\times 10^{-25}$

The another given equilibrium reaction is:

$2NO{(g)}+O_2_{(g)}\rightleftharpoons 2NO_2_{(g)}$

The expression for the equilibrium constant is:

$K_c_2=\frac {[NO_2]^2}{[NO]^2[O_2]}=6.4\times 10^{9}$

To find,

For the equilibrium which is:

$N_2_{(g)}+2O_2_{(g)}\rightleftharpoons 2NO_2_{(g)}$

The expression for the equilibrium constant is:

$K_c=\frac {[NO_2]^2}{[N_2][O_2]^2}$

Multiplying and dividing by $[NO]^2$ and rearranging in the above equation as:

$K_c=\frac {[NO_2]^2}{[NO]^2[O_2]}\times \frac {[NO]^2}{[N_2][O_2]}$

$K_c=K_c_2\times K_c_1=6.4\times 10^{9}\times 4.3\times 10^{-25}=2.752\times 10^{-15}$