mixture of The water-gas reaction is a source of hydrogen. Passing steam over ho

Question

mixture of The water-gas reaction is a source of hydrogen. Passing steam over hot carbon produces a carbon monoxide and hydrogen: CO(g) H The value of Ko for the reaction at 1000 cis 300 x 10-2 a. Calculate the equilibrium partial pressures of the products and reactants if PH20 0.442 atm and Poo 5.000 atm at the start of the reaction. Assume that the carbon is in excess. Number Number PH o .167 at m atm Number atm b. Determine the equilibrium partial pressures of H2o, co, and H2 after co and H2 at 0.079 atm are added to the equilibrium mixture in part a

Explanation / Answer

Recall that

Kp = Kc*(RT)

Kp = (3*10^-2)(0.082*1273) = 3.13158

therefore

a)

Kp = P-CO * P-H2 / (P-H2O)

initially

P-CO = 5

P-H2 = 0.442

P-H2O = 0

in equilibrium

P-CO = 5 - x

P-H2 = 0.442 - x

P-H2O = 0 + x

substitute

Kp = P-CO * P-H2 / (P-H2O)

3.13158 = (5 - x)*(0.442 - x) /(x=

3.13158x = 2.21 - 5.442x +x^2

x^2 -8.55358x + 2.21 = 0

x = 0.266

P-CO = 5 - x 5-0.266 = 4.734

P-H2 = 0.442 - x = 0.442-0.266 = 0.176

P-H2O = 0 + x = 0.266

b)

If we add

initially

P-CO = 4.734 + 0.079 = 4.813

P-H2 = 0.176+0.079 = 0.255

P-H2O = 0.266

in equilibrium

P-CO = 4.734 + 0.079 = 4.813 - x

P-H2 = 0.176+0.079 = 0.255 - x

P-H2O = 0.266 + x

Kp = P-CO * P-H2 / (P-H2O)

3.13158 = (4.813 - x)*(0.255 - x) /(0.266 + x)

3.13158 *0.266 + 3.13158 x = 4.813*0.255 - (0.255 +4.813 )x + x^2

x^2 + (5.068-3.13158)x + 4.813*0.255 - 3.13158 *0.266 = 0

x^2+ 1.93642x + 0.394 = 0

x = -0.23

P-CO = 4.734 + 0.079 = 4.813 - 0.23 = 4.8583

P-H2 = 0.176+0.079 = 0.255 - 0.23 = 0.025

P-H2O = 0.266 +0.23 = 0.496

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