is involved in the formation of smog and acid rain. It is formed importantly as: (g) + (g) ⇌ (g) + ( — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the Reaction Quotient (\(Q_c\)) For the given gas-phase reversible reaction: \[\ce{NO(g) + O3(g) <=> NO2(g) + O2(g)}\] The reaction quotient (\(Q_c\)) at any arbitrary non-equilibrium state is expressed in the same form as the equilibrium constant (\(K_c\)), using the instantaneous molar concentrations of the species: \[Q_c = \frac{[\ce{NO2}][\ce{O2}]}{[\ce{NO}][\ce{O3}]}\] Step 2 - Substitute the Instantaneous Concentrations and Calculate \(Q_c\) We are given the following concentrations in the metropolitan air: * \([\ce{NO}] = 1.0 \times 10^{-5}\text{ M}\) * \([\ce{O3}] = 1.0 \times 10^{-6}\text{ M}\) * \([\ce{NO2}] = 2.5 \times 10^{-4}\text{ M}\) * \([\ce{O2}] = 8.2 \times 10^{-3}\text{ M}\) Substituting these values into our expression for \(Q_c\): \[Q_c = \frac{\left(2.5 \times 10^{-4}\text{ M}\right) \times \left(8.2 \times 10^{-3}\text{ M}\right)}{\left(1.0 \times 10^{-5}\text{ M}\right) \times \left(1.0 \times 10^{-6}\text{ M}\right)}\] Let us compute the numerator and denominator: \[\text{Numerator} = 2.5 \times 8.2 \times 10^{-4 - 3} = 20.5 \times 10^{-7} = 2.05 \times 10^{-6}\text{ M}^2\] \[\text{Denominator} = 1.0 \times 1.0 \times 10^{-5 - 6} = 1.0 \times 10^{-11}\text{ M}^2\] Now, evaluate the ratio: \[Q_c = \frac{2.05 \times 10^{-6}}{1.0 \times 10^{-11}} = 2.05 \times 10^{-6 - (-11)} = 2.05 \times 10^5\] Step 3 - Compare the Reaction Quotient (\(Q_c\)) with the Equilibrium Constant (\(K_c\)) The equilibrium constant for the reaction at the given temperature is: \[K_c = 6.0 \times 10^{34}\] Comparing the two values: \[Q_c = 2.05 \times 10^5 \ll K_c = 6.0 \times 10^{34}\] Step 4 - Determine the Direction of the Reaction Shift According to chemical thermodynamics and Le Chatelier's principle, the relationship between \(Q_c\) and \(K_c\) dictates the direction in which the reaction will shift to attain equilibrium: * If \(Q_c = K_c\), the system is at chemical equilibrium, and no net change in concentrations occurs. * If \(Q_c > K_c\), the reaction shifts in the backward direction to consume products and form reactants. * If \(Q_c < K_c\), the reaction shifts in the forward direction to consume reactants and form products. Since \(Q_c\) (\(2.05 \times 10^5\)) is astronomically smaller than \(K_c\) (\(6.0 \times 10^{34}\)), the system is far from equilibrium and has an immense thermodynamic driving force to shift in the forward direction. Thus, reactants (\(\ce{NO}\) and \(\ce{O3}\)) are consumed, and more of the products (\(\ce{NO2}\) and \(\ce{O2}\)) are formed. Step 5 - Evaluate the Options * **Option (A) "more of \(\ce{NO}\) and \(\ce{O3}\) tend to be formed"**: Incorrect. This would occur if \(Q_c > K_c\), driving the reaction backward. * **Option (B) "more of \(\ce{NO2}\) and \(\ce{O2}\) tend to be formed"**: Correct. Because \(Q_c \ll K_c\), the reaction proceeds strongly in the forward direction, producing more products. * **Option (C) "more of \(\ce{NO2}\) and \(\ce{O3}\) tend to be formed"**: Incorrect. This violates the stoichiometry of the reaction, as one is a product and the other is a reactant. * **Option (D) "no tendency to change because the reaction is at equilibrium"**: Incorrect. The system is not at equilibrium because \(Q_c \neq K_c\). \[\boxed{\text{B}}\]