Please show your work. Question 14.18 The H_2/CO ratio in mixtures of carbon mon

Question

Please show your work. Question 14.18

The H_2/CO ratio in mixtures of carbon monoxide and hydrogen (called synthesis gas) is increased by the water-gas shift reaction CO(g) + H_2 O(g) reversiblearrow CO_2 (g) + H_2 (g). which has an equilibrium constant K_c = 4.24 at 800 K. Calculate the equilibrium concentrations of CO if CO, H_2O, CO_2, and H_2 are added to a reaction vessel with initial concentrations of 0.165 M. Express your answer to three decimal places and include the appropriate units. Calculate the equilibrium concentration H_2O if CO, H_2O, CO_2, AND H_2 are added to a reaction vessel with initial concentrations of 0.165 M. Express your answer to three decimal places and include the appropriate units. Calculate the equilibrium concentration of CO_2 if CO, H_2O, CO_2. and H_2 are added to a reaction vessel with initial concentrations of 0.165 M. Express your answer to three decimal places and include the appropriate units. Calculate the equilibrium concentration of H_2 if CO. H_2O. CO_2. and H_2 are added to a reaction vessel with initial concentrations of 0.165 M. Express your answer to three decimal places and include the appropriate units.

Explanation / Answer

Q = [CO2][H2] / [CO][H2O]
= 0.165*0.165 / (0.165 * 0.165)
= 1

Since Q < k, equilibrium will move to right

CO (g) + H2O (g) <------> CO2 (g) + H2 (g)
0.165 0.165 0.165 0.165
0.165-x 0.165-x 0.165+x 0.165+x

K = (0.165+x)^2 / (0.165-x)^2
4.24 = (0.165+x)^2 / (0.165-x)^2
sqrt (4.24) = (0.165+x) / (0.165-x)
(0.165+x) / (0.165-x) = 2.06
0.165 + x =0.34 - 2.06*x
3.06*x = 0.175
x = 0.057 M

A)
[CO] = 0.165 - x = 0.165 - 0.057 = 0.108 M

B)
[H2O] = 0.165 - x = 0.165 - 0.057 = 0.108 M

C)
[CO2] = 0.165 + x = 0.165 + 0.057 = 0.222 M

D)
[H2] = 0.165 + x = 0.165 + 0.057 = 0.222 M

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