Verified Solution • 2025–26 Session moderate

QJEEM24Ch9ae2: JEE Main 2024 Chemistry - Class 11 JEEM24Ch9ae2 Solution: The following concentrations were observed at for the formation …

The following concentrations were observed at 500 K500 \mathrm{~K} for the formation of NH3\mathrm{NH}_3 from N2\mathrm{N}_2 and H2\mathrm{H}_2 . At equilibrium ; [N2]=2×10−2M,[H2]=3×10−2M\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M} and [NH3]=1.5×10−2M\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M} . Equilibrium constant for the reaction is ________.

Step-by-Step Verified Solution

To calculate the equilibrium constant KC\mathrm{K}_C for the formation of NH3\mathrm{NH}_3 from N2\mathrm{N}_2 and H2\mathrm{H}_2 at 500 K500 \mathrm{~K} , we use the formula:

KC=[NH3]2[N2][H2]3\mathrm{K}_C = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}

Substituting the given concentrations:

KC=(1.5×10−2)2(2×10−2)×(3×10−2)3\mathrm{K}_C = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3}

Calculating the above expression results in:

KC=417\mathrm{K}_C = 417

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