Answer:
0.0243 atm
Explanation:
1. Chemical equation
2PH₃ ⇌ P₂ + 3H₂
E/atm: x 0.417 0.826
2. Kₚ expression
[tex]K_{\text{p}} = \dfrac{p_{\text{P}_{2}}p_{\text{H}_{2}}^{3}}{p_{\text{PH}_{3}}^{2}} = 398[/tex]
Step 3. Calculate the equilibrium concentrations
[tex]\begin{array}{rcl}\\\dfrac{p_{\text{P}_{2}}p_{\text{H}_{2}}^{3}}{p_{\text{PH}_{3}}^{2}}&=&398\\\\\dfrac{0.417 \times 0.826^{3}}{p_{\text{PH}_{3}}^{2}}&=&398\\\\0.2350&=&398 \times{p_{\text{PH}_{3}}^{2}}\\p_{\text{PH}_{3}}^{2}&=& \dfrac{0.2350}{398}\\\\&=&5.905 \times 10^{-4}\\p_{\text{PH}_{3}}&=& \textbf{0.0243 atm}\end{array}\\\text{The equilibrium pressure of PH$_{3}$ is $\large \boxed{\textbf{0.0243} \textbf{ atm}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{0.417 \times 0.826^{3}}{0.0243^{2}}&=&398\\\\398 & = & 398\\\end{array}[/tex]
It checks.