1. Tardigrade
  2. Question
  3. Chemistry
  4. For a reaction X+Y leftharpoons 2 Z, 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L -1. The equilibrium constant of the reaction is (x/15) ⋅ The value of x is.

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Q. For a reaction $X+Y \rightleftharpoons 2 Z, 1.0$ mol of $X, 1.5$ mol of $Y$ and 0.5 mol of $Z$ were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of $Z$ was $1.0 mol L ^{-1}$. The equilibrium constant of the reaction is _____$\frac{x}{15} \cdot$ The value of $x$ is______.

JEE MainJEE Main 2020Equilibrium

Solution:

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$K_{eq} =\frac{1^2}{\frac{3}{4} \times \frac{5}{4}} = \frac{16}{15}$