Question #210965

K= 1.44 at 1000 k for synthesis gas reaction:

CO_{(g)}+H_{2}O →←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?

Expert's answer

CO(g) + H_{2}O(l) "\\leftrightarrow" CO_{2} (g) + H_{2} (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 - 1.44P^{2} = P^{2}

1.44 = 2.44P^{2}

P^{2} = "\\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 CO_{2} (P"_{C0_2}" ) ="\\frac{0.768}{1.768}"

so partial pressure of H_{2} (P"_{H_2}" ) ="\\frac{0.768}{1.768}"

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