JEE MAIN - Chemistry (2024 - 29th January Evening Shift - No. 24)
The following concentrations were observed at $$500 \mathrm{~K}$$ for the formation of $$\mathrm{NH}_3$$ from $$\mathrm{N}_2$$ and $$\mathrm{H}_2$$. At equilibrium ; $$\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M}$$ and $$\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M}$$. Equilibrium constant for the reaction is ________.
Answer
417
Explanation
To calculate the equilibrium constant $\mathrm{K}_C$ for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$ at $500 \mathrm{~K}$, we use the formula:
$ \mathrm{K}_C = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} $
Substituting the given concentrations:
$ \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:
$ \mathrm{K}_C = 417 $
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