Question

$F{e}_2{O}_3\left(s\right)$ may be converted be Fe by the reaction, $F{e}_2{O}_3+3H_2\left(g\right)\rightleftharpoons 2Fe\left(s\right)+3H_{2}O\left(g\right)$; for which $K_p=8$ at temperature ${720}^{0}C.$ What % of $H_2$ remains unreacted after the reaction has come to equilibrium:

• A. $\approx 22$%
• B. $\approx 34$%
• C. $\approx 66$%
• D. $\approx 78$%

Answer 1

The correct option is B $\approx 34$%

Let the initial partial pressure of hydrogen be 100 atm and the initial partial pressure of water be 0 atm.
The equilibrium partial pressures of hydrogen and water will be 100-3P atm and 3 P atm respectively.
The expression for the equilibrium constant is
$\frac{P_{H_{2}O}^3}{P_{H_2}^3}=K_p$
Substitute values in the above expression.
$\frac{(3x{)}^3}{(100-3x{)}^3}=8$
$\frac{3x}{100-3x}=2$
$3x=200-6x$
$9x=200$
$x=22.22$
The percentage of hydrogen that has unreacted after the hydrogen comes to equilibrium is
$\frac{100-3\left(22.22\right)}{100}\times 100=34$ %