Phosphorus pentachloride dissociates on heating:
PCl5(g) ⇄ PCl3(g) + Cl2(g) 𝝙H = 1901 kJ/mol
(a) If Kc equals 3.26 102
at 191°C, what is Kp at this temperature?
(b) State the direction (toward product, toward reactant, no change) that the equilibrium
will shift when each of the following changes occurs. Briefly explain your answers.
PCl5 is added to the system
(i) Pressure is increased
(ii) Volume is increased
(iii) Temperature
(iv) Addition of a catalyst
(v) What change would you expect in the value of equilibrium constant by
increasing the temperature?
pressure squeezes it back left to the fewer mols of gas
volume lets it expand to the right
Temp (heat in) is what you used to make it happen in the first place
To determine the value of Kp at a given temperature, you need to use the equation that relates Kc and Kp.
(a) The equation that relates Kc and Kp is: Kp = Kc(RT)^(∆n), where R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, and ∆n is the change in the number of moles of gas particles (products - reactants).
In this case, there is a decrease in the number of moles of gas particles, as PCl5(g) dissociates into PCl3(g) and Cl2(g). Therefore, ∆n = (1 - 1 - 2) = -2.
Given Kc = 3.26 × 10^(-2) and assuming that the temperature is given in Celsius, you need to convert it to Kelvin. T = 191°C + 273.15 = 464.15 K.
Now you can calculate Kp using the equation:
Kp = Kc(RT)^(∆n)
Kp = (3.26 × 10^(-2))(0.0821 L·atm/(mol·K))(464.15 K)^(-2)
Kp = 1.464 × 10^(-7)
Therefore, Kp at this temperature is approximately 1.464 × 10^(-7).
(b) To determine the direction in which the equilibrium will shift when each change occurs, you can use Le Chatelier's principle.
(i) When PCl5 is added to the system, according to Le Chatelier's principle, the equilibrium will shift toward the products to alleviate the increase in concentration of the reactant.
(ii) When the volume is increased, the equilibrium will shift toward the side with more moles of gas to alleviate the increase in volume. In this case, there are fewer moles of gas on the product side, so the equilibrium will shift toward the products.
(iii) When the temperature is increased, the equilibrium will shift in the endothermic direction to alleviate the increase in temperature. Since the forward reaction is endothermic (𝝙H > 0), the equilibrium will shift toward the products. Conversely, if the temperature is decreased, the equilibrium will shift toward the reactants.
(iv) The addition of a catalyst does not affect the equilibrium position because it speeds up the forward and backward reactions equally. Therefore, there will be no change in the equilibrium position.
(v) Increasing the temperature generally increases the value of the equilibrium constant, as the forward reaction is endothermic. This means that more products will be formed at higher temperatures, resulting in a larger value of Kc.