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.

Question

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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/dcb913331ca79d8bc745cd268bfbe8f1-.png" /> Keq =(12/(3)4 × (5)4 = (16/15)

Q. For a reaction $X + Y ⇌ 2 Z, 1.0$ 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 $1.0 m o l L^{- 1}$ . The equilibrium constant of the reaction is _ $\frac{x}{15} \cdot$ The value of $x$ is__.

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

Q. For a reaction X+Y⇌2Z,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 1.0molL−1. The equilibrium constant of the reaction is _____15x​⋅ The value of x is______.

Keq​=43​×45​12​=1516​

Q. For a reaction $X + Y ⇌ 2 Z, 1.0$ 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 $1.0 m o l L^{- 1}$ . The equilibrium constant of the reaction is _ $\frac{x}{15} \cdot$ The value of $x$ is__.

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