The equilibrium constants $K_p$ and $K_c$ relate to reactions involving gases. $K_p$ is expressed in terms of partial pressures, while $K_c$ is expressed in terms of molar concentrations.
The relationship between $K_p$ and $K_c$ is given by the equation:
$K_p = K_c (RT)^{\Delta n}$
where:
For $K_p$ to be equal to $K_c$, the value of $\Delta n$ must be zero ($K_p = K_c (RT)^0 = K_c$). Therefore, the reaction for which $K_p \neq K_c$ is the one where $\Delta n \neq 0$.
Let's calculate $\Delta n$ for each given reaction:
Reactants: 1 mole $N_2$ + 1 mole $O_2$ = 2 moles of gas
Products: 2 moles $NO$ = 2 moles of gas
$\Delta n = 2 - 2 = 0$. Thus, $K_p = K_c$.
Reactants: 1 mole $H_2O$ + 1 mole $CO$ = 2 moles of gas
Products: 1 mole $H_2$ + 1 mole $CO_2$ = 2 moles of gas
$\Delta n = 2 - 2 = 0$. Thus, $K_p = K_c$.
Reactants: 1 mole $H_2$ + 1 mole $I_2$ = 2 moles of gas
Products: 2 moles $HI$ = 2 moles of gas
$\Delta n = 2 - 2 = 0$. Thus, $K_p = K_c$.
Reactants: 1 mole $N_2$ + 3 moles $H_2$ = 4 moles of gas
Products: 2 moles $NH_3$ = 2 moles of gas
$\Delta n = 2 - 4 = -2$. Since $\Delta n \neq 0$, $K_p \neq K_c$.
The reaction $N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$ is the one for which $K_p \neq K_c$ because the change in the number of moles of gas ($\Delta n$) is not zero.
For a certain reaction R $\rightarrow$ Product, the plot of [R] vs time has a negative slope as shown. The order of reaction is :

| List I (Order of reaction) | List II (Unit of rate constant) |
| A. Zero order | I. $mol^{-1} L s^{-1}$ |
| B. First order | II. $mol^{-2} L^2 s^{-1}$ |
| C. Second order | III. $s^{-1}$ |
| D. Third order | IV. $mol L^{-1} s^{-1}$ |
Calculate emf of the half cell given below :
$$Pt(s) | H_2 (g, 2 \text{ atm}) | HCl (aq, 0.02 \text{ M})$$
$$E_{H_2 /H^+}^\circ = 0 \text{ V}$$
(Given : $\frac{2.303 RT}{F} = 0.059$, $\log 2 = 0.3010$)
At 298 K, a certain buffer solution contains equal concentrations of $X^{-}$ and $HX$. $K_b$ for $X^-$ is $10^{-10}$. What is the pH of this buffer solution ?