A mixture of 0.2000 mol of \(\mathrm{CO}_{2}, 0.1000 \mathrm{mol}\) of \(\mathrm{H}_{2},\) and 0.1600 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000 -L vessel. The following equilibrium is established at \(500 \mathrm{K} :\) $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2,}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}\) \(=3.51\) atm. Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .(\mathbf{c})\) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

Step by step solution

01

Calculate the initial partial pressures of CO2, H2, and H2O.

To calculate the initial partial pressures of the gases, we use the ideal gas law: \( PV = nRT \) Where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Rearranging, we get: \( P = \frac{nRT}{V} \) First, calculate the total pressure: \( P_{total} = \frac{(0.2000~mol + 0.1000~mol + 0.1600~mol)(0.0821~L~atm~mol^{-1}~K^{-1})(500~K)}{2.000~L} \) Next, we'll compute the mole fraction of each gas and multiply it by the total pressure to obtain the initial partial pressure of each gas.

02

Set up the reaction and the equilibrium expression.

We have the following equilibrium reaction: \( CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2}O(g) \) Let's write down the equilibrium expression using partial pressures: \( K_{p} = \frac{P_{CO}P_{H_{2}O}}{P_{CO_{2}}P_{H_{2}}} \)

03

Calculate the equilibrium partial pressures of CO2, H2, and CO.

To calculate the equilibrium partial pressures, we need to relate the initial partial pressures to the changes during the reaction. We are given that the equilibrium partial pressure of H2O is 3.51 atm. Using the stoichiometry of the reaction, we can calculate the amounts of CO2, H2, and CO at equilibrium. Assume that x atm of CO2 and H2 reacted to produce x atm of CO and H2O. Then the equilibrium pressures would be as follows: \(P_{CO_{2}} = P_{CO_{2}}^{initial} - x \) \(P_{H_{2}} = P_{H_{2}}^{initial} - x \) \(P_{CO} = x \) \(P_{H_{2}O} = P_{H_{2}O}^{initial} + x \) We know that the final partial pressure of H2O is 3.51 atm, so we can solve for x and calculate the equilibrium pressures for the other gases.

04

Calculate Kp and Kc for the reaction.

Using the equilibrium partial pressures obtained in step 3, we can calculate the equilibrium constant Kp using the expression: \( K_{p} = \frac{P_{CO}P_{H_{2}O}}{P_{CO_{2}}P_{H_{2}}} \) Now we'll convert Kp to Kc using the relationship between them: \( K_{p} = K_{c}(RT)^{\Delta{n}} \) where \(\Delta{n}\) is the change in the number of moles of gas in the reaction, which is zero in this case. So in this case, Kp = Kc. Finally, we have calculated the initial partial pressures, equilibrium partial pressures, and the equilibrium constants Kp and Kc for the given reaction.

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