Question

\(\mathrm{At} 425^{\circ} \mathrm{C}, K_{\mathrm{p}}=4.18 \times 10^{-9} \mathrm{for}\) the reaction $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g) \pm \mathrm{Br}_{2}(g) $$ In one experiment, 0.20 atm of \(\mathrm{HBr}(g), 0.010\) atm of \(\mathrm{H}_{2}(g),\) and 0.010 atm of \(\mathrm{Br}_{2}(g)\) are introduced into a container. Is the reaction at equilibrium? If not, in which direction will it proceed?

Short answer

The reaction is not at equilibrium and will proceed toward the reactants.

Step-by-step solution

- Write the expression for the equilibrium constant

For the given reaction: $$2 \text{HBr}(g) \rightleftharpoons \text{H}_{2}(g) + \text{Br}_{2}(g)$$ The equilibrium constant in terms of partial pressures (\text{K}_{\text{p}}) is given by: \[ K_{\text{p}} = \frac{P_{\text{H}_2} \cdot P_{\text{Br}_2}}{(P_{\text{HBr}})^2} \] where \(P_{\text{H}_2}\) is the partial pressure of \text{H}_2, \(P_{\text{Br}_2}\) is the partial pressure of \text{Br}_2, and \(P_{\text{HBr}}\) is the partial pressure of \text{HBr}.

- Substitute the given partial pressures

Using the provided partial pressures, \(P_{\text{HBr}} = 0.20 \) atm, \(P_{\text{H}_2} = 0.010 \) atm, and \(P_{\text{Br}_2} = 0.010 \) atm, substitute these values into the \text{K}_{\text{p}} expression: \[ Q_{\text{p}} = \frac{(0.010) \cdot (0.010)}{(0.20)^2} \]

- Calculate the reaction quotient (\text{Q}_{\text{p}})

Calculate \(Q_{\text{p}}\): \[ Q_{\text{p}} = \frac{0.010 \cdot 0.010}{0.20^2} = \frac{0.0001}{0.04} = 0.0025 \]

- Compare \text{Q}_{\text{p}} with \text{K}_{\text{p}}

Given \(K_{\text{p}} = 4.18 \times 10^{-9}\), compare it with \(Q_{\text{p}}\): \[ Q_{\text{p}} (0.0025) \] > \(K_{\text{p}} (4.18 \times 10^{-9}) \)

- Determine direction of reaction

Since \(Q_{\text{p}} > K_{\text{p}}\), the reaction is not at equilibrium. The reaction will proceed in the direction that reduces \(Q_{\text{p}}\), which is to the left (toward the reactants).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\), represents the ratio of the concentrations of products to reactants at equilibrium for a given reaction. For gas-phase reactions, this ratio is often expressed in terms of partial pressures.
This constant can indicate how far a reaction will proceed towards products before reaching equilibrium.
For the given reaction:
\(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g)\),
\(K_p\) is written as:
\[K_p = \frac{P_{\mathrm{H}_2} \cdot P_{\mathrm{Br}_2}}{(P_{\mathrm{HBr}})^2}\]
The value of \(K_p\) does not depend on the initial concentrations of the reactants or products, but it is only valid at a specific temperature.

Reaction Quotient

The reaction quotient, denoted as \(Q_p\), is similar to the equilibrium constant but is calculated at any point during the reaction, not just at equilibrium.
It is a snapshot of the current state of the reaction.
For the reaction:
\(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g)\),
\(Q_p\) is given by:
\[Q_p = \frac{P_{\mathrm{H}_2} \cdot P_{\mathrm{Br}_2}}{(P_{\mathrm{HBr}})^2}\]
By comparing \(Q_p\) to \(K_p\), we can determine the direction the reaction will proceed to reach equilibrium.
- If \(Q_p < K_p\), the reaction will proceed towards products (right).
- If \(Q_p > K_p\), the reaction will proceed towards reactants (left).
- If \(Q_p = K_p\), the reaction is at equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by each gas in a mixture of gases.
Each gas contributes to the total pressure independently of the others.
For our reaction, the partial pressures involved are:
\(P_{\mathrm{HBr}} = 0.20\ \mathrm{atm}\), \(P_{\mathrm{H}_2} = 0.010\ \mathrm{atm}\), and \(P_{\mathrm{Br}_2} = 0.010\ \mathrm{atm}\).
In general, the partial pressure of a gas can be calculated using the ideal gas law, but here they are provided.
Each of these pressures helps determine the reaction quotient (\(Q_p\)) when plugged into the appropriate formula.

Direction of Reaction

The direction in which a reaction proceeds depends on the comparison between \(Q_p\) and \(K_p\).
In our example:\[Q_p = 0.0025\]and\[K_p = 4.18 \times 10^{-9}\]. By comparing these values, we find that \(Q_p > K_p\), meaning the reaction quotient is greater than the equilibrium constant.
This indicates there are more products relative to reactants than at equilibrium.
The reaction will therefore proceed to the left, towards the formation of more reactants (\(\mathrm{HBr}\)) to achieve equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium.
Changes can include variations in concentration, pressure, or temperature.
In our reaction example:
\(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g)\),
If the pressure were increased by adding more \(\mathrm{HBr}\), the system will shift right to lessen the perturbation by producing more \(\mathrm{H}_2\) and \(\mathrm{Br}_2\).
Le Chatelier's Principle helps us predict the qualitative shifts in response to changes, supporting the calculated quantitative results from \(Q_p\) and \(K_p\).