# Problem: When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the following equationCO2(g) + H2(g) ⇌ H2O(g) + CO(g)In one experiment, the following equilibrium concentrations were measured.[H2] = 0.20 mol/L[CO2] = 0.30 mol/L[H2O] = [CO] = 0.55 mol/LWhat is the mole fraction of CO(g) in the equilibrium mixture?Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.Determine Kp in terms of Kc for this system.When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature.

###### FREE Expert Solution

Part 1) We're being asked to calculate the mole fraction of CO(g) in the equilibrium mixture.

Assuming that they all have the same volume (1L):

moles H2 = 0.20 mol
moles CO2 = 0.30 mol
moles H2O = 0.55 mol
moles CO = 0.55 mol

χCO = 0.34

Part 2) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.

91% (267 ratings) ###### Problem Details

When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the following equation

CO2(g) + H2(g) ⇌ H2O(g) + CO(g)

In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L

What is the mole fraction of CO(g) in the equilibrium mixture?

Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction.

Determine Kp in terms of Kc for this system.

When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g).

Calculate the value of Kc at this lower temperature.

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