JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 17)
Fe2N(s) + $${3 \over 2}$$H2(g) ⇌ 2Fe(s) + NH3(g)
Explanation
To determine the correct relationship between the equilibrium constants $K_C$ and $K_p$ for the reaction:
$$\text{Fe}_2\text{N(s)} + \frac{3}{2}\text{H}_2\text{(g)} \rightleftharpoons 2\text{Fe(s)} + \text{NH}_3\text{(g)}$$
First, consider the general relationship between $K_C$ and $K_p$:
$$ K_p = K_C(RT)^{\Delta n} $$
where $\Delta n$ is the change in the number of moles of gas between the products and the reactants, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.
For the given reaction, calculate $\Delta n$:
Moles of gas in products: NH3(g) = 1 mole
Moles of gas in reactants: $${3 \over 2}$$ H2(g) = 1.5 moles
Therefore:
$$ \Delta n = 1 - 1.5 = -0.5 $$
Substitute this value into the equation relating $K_C$ and $K_p$:
$$ K_p = K_C(RT)^{-0.5} $$
Rearrange to express $K_C$:
$$ K_C = K_p(RT)^{0.5} $$
Thus, the correct option is:
Option A: $$K_C = K_p(RT)^{1/2}$$
Comments (0)
