A 1.00-kg sample of Sb2S3(s) and a 10.0-g sample of H2(g) are allowed to react in a 25.0-L container at 713 K. At equilibrium, 72.6 g H2S(g) is present. What is the value of Kp at 713 K for this reaction?
Sb2S3(s) + 3 H2(g) <----> 2Sb(s) + 3 H2S(g)
27,12
To find the value of Kp at 713 K for the given reaction, we need to use the equilibrium concentrations of the reactants and products.
1. First, let's convert the mass of H2S(g) to moles:
- The molar mass of H2S is approximately 34.08 g/mol.
- Therefore, the number of moles of H2S is 72.6 g / 34.08 g/mol = 2.13 mol.
2. Next, let's calculate the mole ratios of the reactants and products. According to the balanced equation, 1 mol of Sb2S3 reacts with 3 mol of H2 to produce 3 mol of H2S. Therefore:
- The number of moles of H2 = 3 * 2.13 mol = 6.39 mol.
- The number of moles of Sb2S3 = 2.13 mol.
- The number of moles of Sb = 2 * 2.13 mol = 4.26 mol.
3. Now, let's calculate the partial pressures of the reactants and products using the ideal gas law:
- The pressure of H2S(g) is given as the equilibrium pressure, so we can use it directly.
- The partial pressure of H2: 6.39 mol / 25.0 L = 0.256 atm (assuming ideal gas behavior).
- The partial pressure of Sb: 4.26 mol / 25.0 L = 0.170 atm (assuming ideal gas behavior).
4. Finally, we can calculate Kp using the formula:
Kp = (PH2S^3) / (PH2^3 * PSb2S3)
Substituting the values into the equation:
Kp = (0.256^3) / (0.170^3 * 1)
Calculating the value:
Kp ≈ 6.8
Therefore, the value of Kp at 713 K for this reaction is approximately 6.8.
To find the value of Kp at 713 K for the given reaction, we need to first set up the expression for the equilibrium constant Kp based on the balanced equation:
Kp = (P_H2S^2)/(P_H2^3)
where P_H2S is the partial pressure of H2S(g) and P_H2 is the partial pressure of H2(g).
To solve for Kp, we need to determine the values of P_H2S and P_H2.
Given:
- Initial mass of Sb2S3 = 1.00 kg (convert to moles using molar mass of Sb2S3 = 339.7 g/mol)
- Initial mass of H2 = 10.0 g (convert to moles using molar mass of H2 = 2.02 g/mol)
- Final mass of H2S = 72.6 g (convert to moles using molar mass of H2S = 34.08 g/mol)
- Volume of the container = 25.0 L
To calculate the partial pressures, we need to use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
For H2(g):
- Calculate the number of moles of H2 using the initial mass (10.0 g) and molar mass of H2 (2.02 g/mol).
- Use the ideal gas law to calculate the partial pressure of H2 at 713 K.
For H2S(g):
- Calculate the number of moles of H2S using the final mass (72.6 g) and molar mass of H2S (34.08 g/mol).
- Use the ideal gas law to calculate the partial pressure of H2S at 713 K.
Finally, substitute the calculated values of P_H2S and P_H2 into the expression for Kp to find the value of Kp at 713 K for the given reaction.