Problem 59 - Chemistry: An Atoms-Focused Approach

Question

If \(K_{c}=1.5 \times 10^{-3}\) for the reaction, $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ in which direction will the reaction proceed if the partial pressures of the three gases are all \(1.00 \times 10^{-3}\) atm?

Step-by-Step Solution

Answer

Answer: The reaction will proceed in the reverse direction, toward the reactants.

Step 1: Calculate Q

Use the given partial pressures of the three gases to calculate the reaction quotient, Q: $$ Q = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} = \frac{(1.00 \times 10^{-3})^2}{(1.00 \times 10^{-3})(1.00 \times 10^{-3})} = 1 $$ Step 2: Compare Q and Kc

Step 2: Compare Q and Kc

Now, compare the calculated Q value (1) with the given Kc value (1.5 × 10^(-3)): $$ Q = 1 > K_{c} = 1.5 \times 10^{-3} $$ Step 3: Determine the direction of the reaction

Step 3: Determine the direction

Since the calculated Q value is greater than the given Kc value, the reaction will proceed in the reverse direction, toward the reactants: $$ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \leftarrow 2 \mathrm{NO}(g) $$

Key Concepts

Understanding the Reaction Quotient

The reaction quotient, often denoted as \(Q\), is used to determine the current state of a chemical reaction relative to its equilibrium state. - It is calculated using the concentration or partial pressures of the reactants and products at any moment in time.- The formula resembles that of the equilibrium constant, \(K\), but it uses the current concentrations or pressures instead of those at equilibrium.By comparing \(Q\) with \(K\), you can predict the direction in which the reaction will proceed:- If \(Q < K\), the reaction will move forward, towards the products.- If \(Q > K\), the reaction will proceed in reverse, towards the reactants.- If \(Q = K\), the system is at equilibrium, and no net change occurs.In the given problem, \(Q\) was found to be 1 and \(K_c\) is 1.5 × 10^{-3}, indicating that \(Q > K_c\), thus the reaction will shift towards the reactants.

Role of Partial Pressure in Chemical Equilibrium

Partial pressure is crucial in reactions involving gases. It refers to the pressure that each gas in a mixture would exert if it occupied the entire volume alone. Partial pressures help to calculate both \(Q\) and \(K_p\) (the equilibrium constant in terms of pressure) for gaseous reactions:- For the reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)\), the partial pressures of \(\mathrm{N}_2\), \(\mathrm{O}_2\), and \(\mathrm{NO}\) are given as 1.00 × 10^{-3} atm.- These values are substituted into the \(Q\) expression to assess the reaction’s direction.Understanding partial pressure helps in visualizing how gases behave under different conditions and influences the prediction of reaction behavior when equilibrium is disturbed.

Exploring the Equilibrium Constant

The equilibrium constant, represented as \(K\), is fundamental in understanding chemical reactions at equilibrium. Different forms of \(K\) exist; \(K_c\) for concentration, and \(K_p\) for partial pressure.- In the context of gas-phase reactions, \(K_p\) is particularly useful since it uses partial pressures.- \(K\) is specific to a particular reaction at a certain temperature, reflecting the ratio of product pressures to reactant pressures when the system is at equilibrium.In the given exercise, the \(K_c\) value is provided as 1.5 × 10^{-3}:- This low \(K_c\) suggests that, at equilibrium, the reactants dominate, meaning very little product is formed.- Comparing \(K\) with \(Q\) helps predict if a reaction will proceed forward, reverse, or remain unchanged once equilibrium is established.Ratio insights from \(K\) facilitate understanding of the extent of reactions, determining conditions for optimization in industrial or laboratory settings.