Question

Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g) ⇌ 2 NO2(g) Kp = ? References: N2(g) + O2(g) ⇌ 2 NO(g) Kp = 2.3 ´ 10–19 ½ N2(g) + O2(g) ⇌ NO2(g) Kp = 8.4 ´ 10–7

Answer 1

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

Best regards.