1 mol of X Y( g ) and 0.2 mol of Y( g ) are mixed in 1 L vessel. At equilibrium, 0.6 mol of Y( g ) is present. The value of K for the reaction X Y( g ) leftharpoons X( g )+Y( g ) is

Question

1 mol of X Y( g ) and 0.2 mol of Y( g ) are mixed in 1 L vessel. At equilibrium, 0.6 mol of Y( g ) is present. The value of K for the reaction X Y( g ) leftharpoons X( g )+Y( g ) is (A) 0.04 mol L -1 (B) 0.06 mol L -1 (C) 0.36 mol L -1 (D) 0.40 mol L -1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/596eb21f71b74f2e43c4c2237891e3a7-.png" /> At eq. [ X ]=0.2+x=0.6 ∴ x=0.4 mol =(0.4/1)=0.4 M [ XY ]=1-0.4 mol =0.6 mol =(0.6/1)=0.6 M [ X ]=x=0.4=(0.4/1)=0.4 M K=([ X ][ Y ]/[ XY ])=(0.4 × 0.6/0.6)=0.4 mol L -1

Q. $1 m o l$ of $X Y (g)$ and $0.2 m o l$ of $Y (g)$ are mixed in $1 L$ vessel. At equilibrium, $0.6 m o l$ of $Y (g)$ is present. The value of $K$ for the reaction
$X Y (g) ⇌ X (g) + Y (g)$ is

$0.04 m o l L^{- 1}$

$0.06 m o l L^{- 1}$

$0.36 m o l L^{- 1}$

$0.40 m o l L^{- 1}$

At eq. $[X] = 0.2 + x = 0.6$
$∴ x = 0.4 m o l = \frac{0.4}{1} = 0.4 M$
$[X Y] = 1 - 0.4 m o l = 0.6 m o l = \frac{0.6}{1} = 0.6 M$
$[X] = x = 0.4 = \frac{0.4}{1} = 0.4 M$
$K = \frac{[ X ] [ Y ]}{[ X Y ]} = \frac{0.4 \times 0.6}{0.6} = 0.4 m o l L^{- 1}$

Q. 1mol of XY(g) and 0.2mol of Y(g) are mixed in 1L vessel. At equilibrium, 0.6mol of Y(g) is present. The value of K for the reaction XY(g)⇌X(g)+Y(g) is

0.04molL−1

0.06molL−1

0.36molL−1

0.40molL−1