Question

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol
of H2O is placed in a 2.000-L vessel. The following equilibrium is
established at 500 K:

CO2(*g*)+H2(*g*)⇌CO(*g*)+H2O(*g*)

Calculate *K**c* for the reaction.

Answer #1

Calculate *Kc* for the reaction.

CO2 (g) + H2 (g) <===> CO (g) + H2O (g)

Kp = (P(CO) x P(H2O)) / (P(CO2) x P(H2))

First we calculate the P of all species as follows:

PV= nRT

R= 0.08206 L atm / K mol, T= 500K V= 2.000L

P(CO2) = nRT / V = 0.2000 mol x 0.08206 L atm / K mol x 500 K /
2.000 L = 4.103 atm

P(H2) = nRT / V = 0.1000 mol x 0.08206 L atm / K mol x 500 K /
2.000 L = 2.052 atm

P(H2O) = nRT / V = 0.1600 mol x 0.08206 L atm / K mol x 500 K /
2.000 L = 3.282 atm

Here n for CO not given means initially CO not formed thus

P(CO) = 0 (initially)

Here equilibrium pressure not given so we cannot calculate the Kp.

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CO2(g)+H2(g)←−→CO(g)+H2O(g)
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.
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1 H2O(g) + 1 CH4(g)--> 1 CO(g) + 3 H2(g)
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Peq(CH4) =
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CO(g)+H2O(g)⇌CO2(g)+H2(g)
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