Question

Question: Consider the reaction between ${\mathrm{H}}_2$and ${\mathrm{O}}_2$at 100 K$K_P=\frac{{\left(P_{{\mathrm{H}}_2\mathrm{O}}\right)}^2}{\left(P_{{\mathrm{O}}_2}\right){\left(P_{{\mathrm{H}}_2}\right)}^2}=1.33\times {10}^{20}$

If 0.500 atm of H₂ and 0.500 atm of O₂are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?

Short answer

The partial pressure of ${\mathrm{H}}_2$$\mathrm{i}\mathrm{s}8.67\times {10}^{-11}\mathrm{M}$

The partial pressure of ${\mathrm{O}}_2$$\mathrm{i}\mathrm{s}0.250\mathrm{M}$

The partial pressure of ${\mathrm{H}}_2\mathrm{O}\mathrm{i}\mathrm{s}0.500\mathrm{M}$

Step-by-step solution

Write equilibrium table.

Given information is

-The equilibrium constant is $K_p=1.33\times {10}^{20}$

-0.50 atm of ${\mathrm{H}}_2$ and 0.500 atm of ${\mathrm{O}}_2$ are allowed to come to equilibrium

We have to find the partial pressures of the components.

We will assume that all ${\mathrm{H}}_2$reacts with ${\mathrm{O}}_2$.

Two moles of ${\mathrm{H}}_2$reacts with 1 mole of ${\mathrm{O}}_2$, to produce 2 moles of ${\mathrm{H}}_2\mathrm{O}$, so after the reaction the partial pressure of ${\mathrm{H}}_2$will be 0 atm (0.500 atm -0.500 atm) of ${\mathrm{O}}_2$ will be 0.250 atm(0.500 atm -0.250 atm, and ${\mathrm{H}}_2\mathrm{O}$ will be 0.500 atm.

Since all of ${\mathrm{H}}_2$ reacted and the partial pressure is 0 , the equilibrium will move to the left

Find the value of x.

$K_p=\frac{{\left(H_2\mathrm{O}\right)}^2}{{\left(H_2\right)}^2\times \left(O_2\right)}$

$1.33\cdot {10}^{20}=\frac{{(0.500-x)}^2}{{(2x)}^2\times (0.250+x)}$

Since ${\mathrm{K}}_{-}p$ is larger than ${10}^4$

we will assume that 0.500 $-\mathbf{X}\approx$0.500

and that 0.250$+x\approx 0.250$

$1.33\times {10}^{20}=\frac{{(0.500)}^2}{{(2x)}^2\times (0.250)}$

$4x^2=\frac{0.230}{1.33\times {10}^{20}\times 0.250}$

$x^2=1.88\times {10}^{-21}$

$x=4.34\times {10}^{-11}\mathrm{M}$

Therefore, the partial pressures of the components, are

$\begin{array}{c}{P}_{H_2}=2x=8.67\times {10}^{-11}\mathrm{M}\\ P_{O_2}=0.250+x=0.250\mathrm{M}\\ P_{H_{2}O}=0.500-x=0.500\mathrm{M}\end{array}$