Question

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) Dissociation of}{N}_2{O}_4\left(g\right):& \text{(P) Increases with}\\ N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right);& \text{temperature}\\ \Delta H=+57.0kJ& \\ \text{(II) Oxidation of}N{H}_3\left(g\right):& \text{(Q) Decreases with}\\ 4N{H}_3\left(g\right)+5O_2\rightleftharpoons 4NO\left(g\right)+6H_{2}O\left(g\right);& \text{pressure}\\ \Delta H=-900.0kJ& \\ \text{(III) Oxidation of nitrogen:}& \text{(R) Increases with}\\ N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right);& \text{addition of inert}\\ \Delta H=+180.0kJ& \text{gas at constant}\\ & \text{pressure}\\ \text{(IV) Formation of}N{O}_2\left(g\right):& \text{(S) Decreases with}\\ NO\left(g\right)+O_3\left(g\right)\rightleftharpoons N{O}_2\left(g\right)+O_2\left(g\right);& \text{temperature}\\ \Delta H=-200.0kJ& \\ & \text{(T)}{K}_p=K_c\\ & \text{(U)}{K}_p=K_c(RT{)}^1\end{array}$

Which of the following options has the correct combination considering List-I and List-II?

• A. $III\to P,T;IV\to S,U$
• B. $III\to U;IV\to S,T$
• C. $III\to P,T;IV\to R,S,T$
• D. $III\to P,T;IV\to S,T$

Answer 1

The correct option is D $III\to P,T;IV\to S,T$

$III$. It is an endothermic reaction therefore on increasing the temperature, reaction will proceed in the forward direction.
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$
$K_p=K_c$, since $\Delta n_g=0$

$IV$. It is an exothermic reaction therefore on increasing the temperature, reaction moves in the backward direction.
$NO\left(g\right)+O_3\left(g\right)\rightleftharpoons N{O}_2\left(g\right)+O_2\left(g\right)$
$K_p=K_c$, since $\Delta n_g=0$