Question

K= 1.44 at 1000 k for synthesis gas reaction:

CO_(g)+H₂O →←CO_2(g)+H_2(g)

if initially the gases are introduced into an empty vessel at a partial pressure at 1.0 ATM each, what will be the equilibrium partial pressure of each gas?

Answer 1

CO(g) + H₂O(l) $\leftrightarrow$ CO₂ (g) + H₂ (g)

t=0 .... 1........--...........--...........--

.......1-P.....................P...........P

total pressure at equilibrium = 1+P

partial pressure of P_CO = $\frac{1-P}{1+P}$ at equilibrium

partial pressure of P$_{CO_2}$ = $\frac{P}{1+P}$ at equilibrium

partial pressure of P$_{H_2}$ = $\frac{P}{1+P}$ at equilibrium

k = $\frac{(\frac{P}{1+P})(\frac{P}{1+P})}{(\frac{1-P}{1+P})}$ = $\frac{P^2}{1-P^2}$

1.44 = $\frac{P^2}{1-P^2}$

1.44 - 1.44P² = P²

1.44 = 2.44P²

P² = $\frac{1.44}{2.44}$

P = 0.768

so partial pressure of CO (P_CO) =$\frac{1-0.768}{1+0.768}$

so partial pressure of CO₂ (P$_{C0_2}$ ) =$\frac{0.768}{1.768}$

so partial pressure of H₂ (P$_{H_2}$ ) =$\frac{0.768}{1.768}$