At 550 K, the Kc for the following reaction is 104 mol -1 L X(g)+Y(g) leftharpoons Z(g) At equilibrium, it was observed that [X]=(1/2)[Y]=(1/2)[Z] What is the value of [Z] (in mol L -1 ) at equilibrium?

Question

At 550 K, the Kc for the following reaction is 104 mol -1 L X(g)+Y(g) leftharpoons Z(g) At equilibrium, it was observed that [X]=(1/2)[Y]=(1/2)[Z] What is the value of [Z] (in mol L -1 ) at equilibrium? (A) 2 × 10-4 (B) 10-4 (C) 2 × 104 (D) 104

Tardigrade Admin · Accepted Answer

Let [X]=(1/2)[Y]=(1/2)[Z]=x then [Y]=2 x and [Z]=2 x,[X]=x Now, K=([Z]/[X][Y]) ⇒ 104=(2 x/x × 2 x) ⇒ x=10-4 ∴ [Z]=2 x=2 × 10-4

Q. At $550 K$ , the $K_{c}$ for the following reaction is $1 0^{4} m o l^{- 1} L$
$X_{(g)} + Y_{(g)} ⇌ Z_{(g)}$
At equilibrium, it was observed that
$[X] = \frac{1}{2} [Y] = \frac{1}{2} [Z]$
What is the value of $[Z]$ (in $m o l L^{- 1}$ ) at equilibrium?

$2 \times 1 0^{- 4}$

$1 0^{- 4}$

$2 \times 1 0^{4}$

$1 0^{4}$

Let $[X] = \frac{1}{2} [Y] = \frac{1}{2} [Z] = x$

then $[Y] = 2 x$ and $[Z] = 2 x, [X] = x$

Now, $K = \frac{[ Z ]}{[ X ] [ Y ]}$

$\Rightarrow 1 0^{4} = \frac{2 x}{x \times 2 x}$

$\Rightarrow x = 1 0^{- 4}$

$∴ [Z] = 2 x = 2 \times 1 0^{- 4}$

Q. At 550K, the Kc​ for the following reaction is 104mol−1L X(g)​+Y(g)​⇌Z(g)​ At equilibrium, it was observed that [X]=21​[Y]=21​[Z] What is the value of [Z] (in molL−1 ) at equilibrium?

2×10−4

10−4

2×104

104