An industrial chemist puts 1.00 mol each of H2(g) and CO2(g) in a 1.00-L container at 800oC. When equilibrium is established, 0.49 mol of CO(g) is in the container. Find Kc at 800 C for the following reaction:
H2(g) + CO2(g) ⇌ H2O(g) + CO(g).
H2 + CO2 <> H2O + C0 all gases.
K= [H2O][CO]/[H2}{CO2]
K= (.49)^2 /(1-.49)(1-.49)=(.49/.51)^2
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To find Kc at 800°C for the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g), we need to use the given information.
First, let's write down the balanced chemical equation for the reaction:
H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
The stoichiometric coefficients in the balanced chemical equation give us the ratio of moles of reactants and products.
According to the given information, the initial moles of H2(g) and CO2(g) are both 1.00 mol. After the reaction reaches equilibrium, there are 0.49 mol of CO(g) present.
To find the equilibrium concentrations, we need to consider the stoichiometry of the reaction. The equilibrium concentrations can be represented as:
[H2] = 1.00 - x
[CO2] = 1.00 - x
[H2O] = x
[CO] = 0.49 + x
Here, x represents the change in moles from the initial moles.
Now, let's substitute the equilibrium concentrations into the equilibrium expression for Kc:
Kc = [H2O][CO] / [H2][CO2]
Substituting the equilibrium concentrations, we get:
Kc = (x)(0.49 + x) / (1.00 - x)(1.00 - x)
Since we know the moles of CO(g) at equilibrium (0.49 mol), we can substitute that into the equation:
Kc = (x)(0.49 + x) / (1.00 - x)(1.00 - x) = (0.49 + x) / (1.00 - x)
Now, to solve for x, we use the fact that the moles of CO(g) at equilibrium is 0.49 mol:
0.49 = (0.49 + x) / (1.00 - x)
Cross-multiplying and simplifying, we have:
0.49(1 - x) = 0.49 + x
0.49 - 0.49x = 0.49 + x
Combining like terms, we get:
1.98x = 0
Dividing both sides by 1.98, we find:
x = 0
Since x = 0, the change in moles from the initial moles is zero, which means the reaction is at equilibrium.
Plugging in the value of x into the equilibrium expression for Kc, we get:
Kc = (0.49 + 0) / (1.00 - 0) = 0.49
Therefore, Kc at 800°C for the reaction H2(g) + CO2(g) ⇌ H2O(g) + CO(g) is 0.49.