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Question: Answered & Verified by Expert
At $550 \mathrm{~K}$, the $K_c$ for the following reaction is $10^4 \mathrm{~mol}^{-1} \mathrm{~L} X(g)+Y(g) \rightleftharpoons Z(g)$

At equilibrium, it was observed that $[X]=\frac{1}{2}[Y]=\frac{1}{2}[Z]$

What is the value of [Z] (in $\mathrm{mol} \mathrm{L}^{-1}$ ) at equilibrium?
ChemistryChemical EquilibriumJEE Main
Options:
  • A $2 \times 10^{-4}$
  • B $10^{-4}$
  • C $2 \times 10^4$
  • D $10^4$
Solution:
1548 Upvotes Verified Answer
The correct answer is: $2 \times 10^{-4}$
$\begin{aligned} & X(g)+Y(g) \rightleftharpoons Z(g) \\ & K_c=\frac{[Z]}{[X][Y]} \text { or } 10^4=\frac{[Z]}{[X][Y]} \\ & \therefore \quad[Z]=10^4[X][Y]=10^4 \times \frac{1}{2}[Z][Z] \\ & \text { (Given, }[X]=\frac{1}{3}[Y]=\frac{1}{2}[Z] \text { ) } \\ & \therefore \quad[Z]=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \\ & \end{aligned}$

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