JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
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 ________.
- A \(419\)
- B \(418\)
- C \(417\)
- D \(455\)
Answer & Solution
Correct Answer
(C) \(417\)
Step-by-step Solution
Detailed explanation
\( \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \) \( \mathrm{~K}_{\mathrm{C}}=\frac{\left(1.5 \times 10^{-2}\right)^2}{\left(2 \times 10^{-2}\right) \times\left(3 \times 10^{-2}\right)^3} \)…
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