Question

The equilibrium constant for the decomposition of water,
$H_{2}O\left(g\right)\to H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
is given by:

• A. $K_p=\frac{{\alpha }^2{p}^{\frac{1}{2}}}{(1+\alpha )(2-\alpha {)}^{\frac{1}{2}}}$
• B. $K_p=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{(1-\alpha )(2+\alpha {)}^{\frac{1}{2}}}$
• C. $K_p=\frac{{\alpha }^3{p}^{\frac{1}{2}}}{\sqrt{2}}$
• D. $K_p=\frac{{\alpha }^3{p}^{\frac{1}{2}}}{(1-\alpha )(2+\alpha {)}^{\frac{1}{2}}}$

Answer 1

The correct option is B $K_p=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{(1-\alpha )(2+\alpha {)}^{\frac{1}{2}}}$

$\begin{array}{cccccc}& H_{2}O\left(g\right)& \to & H_2\left(g\right)& +& \frac{1}{2}{O}_2\left(g\right)\\ t=0& 1& & 0& & 0\\ t=eq& 1-\alpha & & \alpha & & \frac{\alpha }{2}\end{array}$
So $K_p=\frac{P_{H_2}(P_{O_2}{)}^{\frac{1}{2}}}{P_{H_{2}O}}$
Total moles $=1-\alpha +\alpha +\frac{\alpha }{2}=1+\frac{\alpha }{2}$
$P_{H_2}=\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right)\times P$
$P_{H_{2}O}=\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right).P$
$P_{O_2}=\left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right).P$
$K_p=\frac{\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right).P{(\left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right).P)}^{\frac{1}{2}}}{(\frac{1-\alpha }{1+\frac{\alpha }{2}}.P)}$ $=\frac{{\alpha }^{\frac{3}{2}}.P^{\frac{1}{2}}}{(1-\alpha )(2+\alpha {)}^{\frac{1}{2}}}$