Question

Assume that the decomposition of $HNO _{3}$ can be represented by the following equation
$4 HNO _{3}( g ) \rightleftharpoons 4 NO _{2}( g )+2 H _{2} O ( g )+ O _{2}( g )$
and the reaction approaches equilibrium at $400\, K$ temperature and $30\, atm$ pressure. At equilibrium partial pressure of $HNO _{3}$ is $2\, atm .$ Calculate $K_{C}$ in $( mol / L )^{3}$ at $400\, K$
(Use: $R=0.08\, atm\, \cdot L /\, mol \cdot K )$

• A. 4
• B. 8
• C. 16
• D. 32

Answer 1

Answer: (D)

$P_{\text {total}}= P _{ HNO _{3}}+ P _{ NO _{2}}+ P _{ H _{2} O }+ P _{ O _{2}}$

$\therefore P _{ NO _{2}}=4 P _{ O _{2}}$ and $P _{ H _{2} O }=2 P _{ O 2}$

$\because P_{\text {total }}= P _{ HNO _{3}}+7 P _{ O _{2}}$

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

$\Rightarrow P_{ O _{2}}=\frac{28}{7}=4$

$K_{P}=\frac{ P _{ NO _{2}}^{4} \cdot P _{ H _{2} O} \cdot P _{ O _{2}}}{ P _{ HNO _{3}}^{4}}$

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

$K_{P} =K_{C}( RT )^{-\Delta n g}=K_{C}(0.08 \times 400)^{3}$where $\Delta n g=3$

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