TS EAMCET · Chemistry · Chemical Equilibrium
For the formation of \(\mathrm{NH}_3\) from \(\mathrm{N}_2\) and \(\mathrm{H}_2\) at \(500 \mathrm{~K}\), the concentration of \(\mathrm{N}_2, \mathrm{H}_2\) and \(\mathrm{NH}_3\) at equilibrium are \(1.5 \times 10^{-2} \mathrm{M}, 3.0 \times 10^{-2} \mathrm{M}\) and \(1.2 \times 10^{-2} \mathrm{M}\), respectively. The equilibrium constant for the reverse reaction is
- A \(3.56 \times 10^2\)
- B \(2.81 \times 10^{-3}\)
- C \(3.56 \times 10^{-2}\)
- D \(2.81 \times 10^3\)
Answer & Solution
Correct Answer
(B) \(2.81 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
Given: Concentration of \(\begin{aligned} {\left[\mathrm{N}_2\right] } & =1.5 \times 10^{-2} \mathrm{M} \\ {\left[\mathrm{H}_2\right] } & =3.0 \times 10^{-2} \mathrm{M} \\ {\left[\mathrm{NH}_3\right] } & =1.2 \times 10^{-2} \mathrm{M} \end{aligned}\) The reaction for the…
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