Question

At \(460^{\circ} \mathrm{C}\), the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g) $$ has \(K=84.7\). All gases are at an initial pressure of 1.25 atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

Short answer

Answer: The partial pressures of each gas at equilibrium are: SO₂ = 0.178 atm, NO₂ = 0.178 atm, NO = 2.322 atm, and SO₃ = 2.322 atm. The total pressure initially is 2.50 atm, and at equilibrium, it increases to 5.00 atm.

Step-by-step solution

Introduce Variables for Equilibrium Pressures

Denote the equilibrium partial pressures as follows: - \(P_{SO_2}\) for \(\mathrm{SO_2}\) at equilibrium - \(P_{NO_2}\) for \(\mathrm{NO_2}\) at equilibrium - \(P_{NO}\) for \(\mathrm{NO}\) at equilibrium - \(P_{SO_3}\) for \(\mathrm{SO_3}\) at equilibrium

Calculate Change in Pressure for the Reaction

Since the reaction has a 1 : 1 stoichiometry, the change in pressure for each species will be the same. Let this change equal x. Then, we can write the following expressions for all pressures at equilibrium: - \(P_{SO_2} = 1.25 - x\) - \(P_{NO_2} = 1.25 - x\) - \(P_{NO} = 1.25 + x\) - \(P_{SO_3} = 1.25 + x\)

Write the Expression for Reaction Quotient Q

Write the reaction quotient Q using the equilibrium constant (K) and the equilibrium partial pressures (\(P_{NO}\), \(P_{SO_3}\), \(P_{SO_2}\), and \(P_{NO_2}\)): \(Q = \frac{P_{NO} \cdot P_{SO_3}}{P_{SO_2} \cdot P_{NO_2}}\)

Substitute Pressures at Equilibrium in Q Equation

Substitute the expressions for equilibrium partial pressures from Step 2 into the reaction quotient equation: \(Q = \frac{(1.25 + x) \cdot (1.25 + x)}{(1.25 - x) \cdot (1.25 - x)}\) Now, we know that Q = K at equilibrium, which is equal to 84.7. So, we can set up the equation as follows: \(84.7 = \frac{(1.25 + x) \cdot (1.25 + x)}{(1.25 - x) \cdot (1.25 - x)}\)

Solve for x

Solve the equation above for x. You will find that x = 1.072 atm.

Calculate the Partial Pressures at Equilibrium

Substitute the value of x back into the expressions from Step 2: - \(P_{SO_2} = 1.25 - 1.072 = 0.178\) atm - \(P_{NO_2} = 1.25 - 1.072 = 0.178\) atm - \(P_{NO} = 1.25 + 1.072 = 2.322\) atm - \(P_{SO_3} = 1.25 + 1.072 = 2.322\) atm These are the partial pressures of each gas at equilibrium.

Compare the Total Pressure Initially and at Equilibrium

Initially, the total pressure is the sum of the initial pressures: \(P_{initial} = 1.25 + 1.25 = 2.50\) atm At equilibrium, the total pressure is the sum of the equilibrium pressures: \(P_{equilibrium} = 0.178 + 0.178 + 2.322 + 2.322 = 5.00\) atm We see that the total pressure has doubled when moving from the initial state to the equilibrium state. However, this does not hold true for all gaseous systems. The change in total pressure depends on the stoichiometry of each reaction and the initial conditions.