Let the degree of decomposition be $\alpha$

Start with 1 mole of steam: $\mathrm{H_2O(g) \rightleftharpoons H_2(g) + \tfrac12 O_2(g)}$

At equilibrium:

• Moles of $\mathrm{H_2O}$ = $1-\alpha$
• Moles of $\mathrm{H_2}$ = $\alpha$
• Moles of $\mathrm{O_2}$ = $\dfrac{\alpha}{2}$

Total moles at equilibrium

$n_{\text{total}}=(1-\alpha)+\alpha+\frac{\alpha}{2}=1+\frac{\alpha}{2}$

Mole fractions

$y_{\mathrm{H_2O}}=\frac{1-\alpha}{1+\alpha/2},\qquad y_{\mathrm{H_2}}=\frac{\alpha}{1+\alpha/2},\qquad y_{\mathrm{O_2}}=\frac{\alpha/2}{1+\alpha/2}$

Use equilibrium constant in terms of partial pressures

Given: $K_p=2\times10^{-3}$

Since total pressure $P=1\,$bar, $p_i=y_iP=y_i$ numerically (in bar standard state treatment).

Therefore, $K_p=\frac{p_{\mathrm{H_2}}\,p_{\mathrm{O_2}}^{1/2}}{p_{\mathrm{H_2O}}}$

Substituting mole fractions: $K_p=\frac{\dfrac{\alpha}{1+\alpha/2}\left(\dfrac{\alpha/2}{1+\alpha/2}\right)^{1/2}}{\dfrac{1-\alpha}{1+\alpha/2}}$

The factor $\left(1+\alpha/2\right)$ cancels partially: $K_p=\frac{\alpha\sqrt{\alpha/2}}{(1-\alpha)\sqrt{1+\alpha/2}}$

Since $K_p$ is very small, $\alpha$ will be small

So approximately: $1-\alpha\approx 1,\qquad 1+\frac{\alpha}{2}\approx 1$

Hence, $K_p\approx \alpha\sqrt{\frac{\alpha}{2}}=\frac{\alpha^{3/2}}{\sqrt{2}}$

Put $K_p=2\times10^{-3}$: $2\times10^{-3}=\frac{\alpha^{3/2}}{\sqrt{2}}$

$\alpha^{3/2}=2\times10^{-3}\times\sqrt{2}$

$\alpha^{3/2}\approx 2\times10^{-3}\times1.414=2.828\times10^{-3}$

Solve for $\alpha$

$\alpha=(2.828\times10^{-3})^{2/3}$

Noting that $2.828\times10^{-3}=2\sqrt{2}\times10^{-3}$ gives $\alpha\approx 0.02$

So the fraction decomposed is about $0.02$.

Convert to percent decomposition

$\%\text{ decomposition}=\alpha\times100=0.02\times100=2\%$

Nearest integer

$\boxed{2}$