Question

Assume that the decomposition of $HN{O}_3$ can be represented by the following equation
$4HN{O}_3(g)\rightleftharpoons 4N{O}_2(g)+2H_{2}O(g)+O_2(g)$
and the reaction approaches equilibrium at $400K$ temperature and $30atm$ pressure. At equilibrium partial pressure of $HN{O}_3$ is $2atm.$ Calculate $K_C$ in $(mol/L{)}^3$ at $400K$
(Use: $R=0.08atm\cdot L/mol\cdot K)$

• A. 4
• B. 8
• C. 16
• D. 32

Answer 1

Answer: (D)

$P_{\text{total}}=P_{HN{O}_3}+P_{N{O}_2}+P_{H_{2}O}+P_{O_2}$

$\therefore P_{N{O}_2}=4P_{O_2}$ and $P_{H_{2}O}=2P_{O2}$

$\because P_{\text{total }}=P_{HN{O}_3}+7P_{O_2}$

$\Rightarrow 30-2=P_{O_2}\times 7$

$\Rightarrow P_{O_2}=\frac{28}{7}=4$

$K_P=\frac{P_{N{O}_2}^4\cdot P_{H_{2}O}\cdot P_{O_2}}{P_{HN{O}_3}^4}$

$=\frac{(4\times 4{)}^4\times (2\times 4{)}^2\times 4}{2^4}=2^{20}$

$K_P=K_C(RT{)}^{-\Delta ng}=K_C(0.08\times 400{)}^3$where $\Delta ng=3$

$\Rightarrow K_C=\frac{2^{20}}{(32{)}^3}=32$