Question

When $0.5mol$ of $N_2{O}_4$ is placed in a $4.00L$ reaction vessel and heated at $400K$, $79.3$ % of the $N_2{O}_4$ decomposes to $N{O}_2$. Calculate $K_c$ and $K_p$ at $400K$ for the reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ :

• A. $K_c=1.51,K_p=49.6$
• B. $K_p=1.51,K_c=49.6$
• C. $K_c=3.0,K_p=99.2$
• D. None of these

Answer 1

The correct option is A $K_c=1.51,K_p=49.6$

Initial number of moles of $N_2{O}_4$ are $0.5$ mol
Out of this $79.3$ % decompose.

Moles of $N_2{O}_4$ decomposed $=0.5\times \frac{79.3}{100}=0.3965$ moles

The number of moles of $N_2{O}_4$ that remains at equilibrium are $0.5-0.3965=0.1035$ moles

The number of moles of $N{O}_2$ formed at equilibrium are $2\times 0.3965=0.793$

The equilibrium constant expression is $K_c=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}=\frac{(\frac{0.793}{4}{)}^2}{\frac{0.1035}{4}}=1.51$

The relationship between Kp and Kc is $K_p=K_c(RT{)}^{\Delta n}$

For the given reaction, $\Delta n=2-1=0$

Hence, $K_p=1.51(0.08206\times 400{)}^1=49.6$