AP EAMCET · Chemistry · Chemical Equilibrium
For the gaseous reactions (I) and (II), the equilibrium constants are \(X\) and \(Y\) respectively.
I. \(\frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)\)
II. \(2 \mathrm{NO}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(g)\)
Using the above reactions the equilibrium constant \(Z\) for the reaction (III) given below is
III. \(\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons \mathrm{N}_2(g)+2 \mathrm{O}_2(g)\)
- A \(Z=X Y\)
- B \(Z=\frac{Y^2}{X}\)
- C \(Z=\frac{1}{X Y^2}\)
- D \(Z=\frac{1}{X^2 Y}\)
Answer & Solution
Correct Answer
(D) \(Z=\frac{1}{X^2 Y}\)
Step-by-step Solution
Detailed explanation
For reaction \(\begin{aligned} \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g) & \rightleftharpoons \mathrm{NO}_2(g) \\ X & =K_{C_1}=\frac{\left[\mathrm{NO}_2\right]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]} \end{aligned}\)…
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