Question

The equilibrium constant for the reaction is $H_2+B{r}_2\to 2HBr$ at 1024 k is $1.60\times {10}^5.$ Find the equilibrium pressure of all gases if $10bar$ of HBr is introduced into a sealed container at $1024K$.

• A. $10bar$
• B. $20bar$
• C. $15bar$
• D. $17bar$

Answer 1

The correct option is A $10bar$

$H_2+B{r}_2\iff 2HBr;Kp=1.6\times {10}^{5}at1024K$
when reaction will be reverse,
$2HBr\iff H_2+B{r}_2;K^{\prime }p=1/K_p=1/1.6\times {10}^{5}at1024K$

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.

$K^{\prime }p=P\left(H_2\right)\times P\left(B{r}_2\right)/P^2\left(HBr\right)$
$=\left(x/2\right).\left(x/2\right)/(10-x{)}^2$
$1/1.6\times {10}^5=x^2/4(10-x{)}^2$
$0.625\times {10}^5=x^2/4(10-x{)}^2$
because values of K'p is so small so, $(10-x)\approx 10$
$0.625\times {10}^{-5}=x^2/4\times 100$
$625\times 4\times {10}^{-5}=x^2$
$x=0.05$
$x/2=0.025=2.5\times {10}^{-2}bar$
Hence, $P\left(H_2\right)=P\left(B{r}_2\right)=2.5\times {10}^{-2}bar$
so, $P\left(HBr\right)=10-x=10-0.05\approx 10bar$