JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 17)

For the reaction

Fe2N(s) + $${3 \over 2}$$H2(g) ⇌ 2Fe(s) + NH3(g)
KC = Kp(RT)1/2
KC = Kp(RT)-1/2
KC = Kp(RT)
KC = Kp(RT)3/2

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: $$K_C = K_p(RT)^{1/2}$$

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