The equilibrium constant for the reaction is H2 - Br2 - 2HBr at 1024 k is 1-60 - 105- Find the equilibrium pressure of all gases if 10 bar of HBr is introduced into a sealed container at 1024 K-

H_2 + Br_2 ⇔ 2HBr; Kp = 1.6 × 10^5 at 1024K when reaction will be reverse, nbsp; 2HBr ⇔ H_2 + Br_2; K'p = 1/K_p = 1/1.6 × 10^5 at 1024 K at initial time, pr ...

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Standard XIII

Chemistry

Characteristics of Chemical Equilibrium

The equilibri...

Question

The equilibrium constant for the reaction is $H_{2} + B r_{2} \to 2 H B r$ at 1024 k is $1.60 \times 10^{5} .$ Find the equilibrium pressure of all gases if $10 b a r$ of HBr is introduced into a sealed container at $1024 K$ .

10 b a r

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Solution

The correct option is A $10 b a r$
$H_{2} + B r_{2} \Leftrightarrow 2 H B r; K p = 1.6 \times 10^{5} a t 1024 K$
when reaction will be reverse,
$2 H B r \Leftrightarrow H_{2} + B r_{2}; K^{'} p = 1 / K_{p} = 1 / 1.6 \times 10^{5} a t 1024 K$

at initial time, pressure of HBr is 10 bar pressure of $H_{2}$ and $B r_{2}$ are 0 bar.

at equilibrium, pressure of HBr is (10 - x) bar pressure of $H_{2}$ and $B r_{2}$ are x/2 bar.

$K^{'} p = P (H_{2}) \times P (B r_{2}) / P^{2} (H B r)$
$= (x / 2) . (x / 2) / (10 - x)^{2}$
$1 / 1.6 \times 10^{5} = x^{2} / 4 (10 - x)^{2}$
$0.625 \times 10^{5} = x^{2} / 4 (10 - x)^{2}$
because values of K'p is so small so, $(10 - x) \approx 10$
$0.625 \times 10^{- 5} = x^{2} / 4 \times 100$
$625 \times 4 \times 10^{- 5} = x^{2}$
$x = 0.05$
$x / 2 = 0.025 = 2.5 \times 10^{- 2} b a r$
Hence, $P (H_{2}) = P (B r_{2}) = 2.5 \times 10^{- 2} b a r$
so, $P (H B r) = 10 - x = 10 - 0.05 \approx 10 b a r$

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The equilibrium constant for the reaction is H2+Br2→2HBr at 1024 k is 1.60×105. Find the equilibrium pressure of all gases if 10 bar of HBr is introduced into a sealed container at 1024 K.

The equilibrium constant for the reaction is $H_{2} + B r_{2} \to 2 H B r$ at 1024 k is $1.60 \times 10^{5} .$ Find the equilibrium pressure of all gases if $10 b a r$ of HBr is introduced into a sealed container at $1024 K$ .