Question

Given that \(K_{c}=62\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 . \mathrm{K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \mathrm{mmol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 1.24 \mathrm{mmol} \cdot \mathrm{L}^{-1}\) \(\mathrm{H}_{2}\), and \(0.112 \mathrm{mmol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\) is present in a container at 500 . \(\mathrm{K}\).

Short answer

More ammonia will form as the reaction quotient (Q = 0.002952) is less than the equilibrium constant (Kc = 62).

Step-by-step solution

Determining the Reaction Quotient (Q)

To decide if more ammonia will form, we need to calculate the reaction quotient (Q) and compare it to the equilibrium constant (Kc). The reaction quotient is calculated by the formula: Q = \( \frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3} \), where [NH_3], [N_2], and [H_2] are the molar concentrations of ammonia, nitrogen, and hydrogen, respectively, at non-equilibrium conditions.

Substituting Given Concentrations

Substitute the given concentrations into the Q expression: \( Q = \frac{(0.112)^2}{(2.23)(1.24)^3} \).

Calculating the Reaction Quotient (Q)

Calculate Q using the substituted values: \( Q = \frac{(0.112)^2}{(2.23)(1.24)^3} \approx \frac{0.012544}{(2.23)(1.906496)} \approx \frac{0.012544}{4.25038368} \approx 0.002952 \).

Comparing Q to Kc

Compare the calculated Q to the given Kc value. If Q is less than Kc, the reaction will proceed in the forward direction, forming more NH3. If Q is equal to Kc, the system is at equilibrium, and if Q is greater than Kc, the reaction will proceed in the reverse direction, forming more N2 and H2.

Determining the Direction of the Reaction

Since Q (0.002952) is less than Kc (62), the reaction will proceed in the forward direction, and more ammonia (NH3) will be formed until equilibrium is reached.

Key concepts

Chemical Equilibrium and the Reaction Quotient (Q)

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. At this state, the amounts of reactants and products remain constant. To predict the direction in which a reaction mixture will proceed to reach equilibrium, we use the reaction quotient (Q).

Q represents the ratio of the product of concentrations of reaction products to the product of concentrations of reactants, each raised to the power of their respective coefficients in the balanced chemical equation. Importantly, only species in the gaseous or aqueous phase are included in the Q expression. For a generic reaction where `aA + bB rightleftharpoons cC + dD`, the reaction quotient is calculated as: \[ Q = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b} \].

In the given exercise, Q is determined for the synthesis of ammonia from nitrogen and hydrogen gas: \[ Q = \frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3} \].
To find Q, you plug in the molar concentrations of each species observed before the system has reached equilibrium. By comparing Q to the equilibrium constant (Kc), we can predict the direction the reaction will shift to achieve equilibrium. If \( Q < Kc \), the forward reaction is favored, and more product is expected to form.

Understanding the Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a value that reflects the ratio of the concentrations of products to reactants at chemical equilibrium for a reaction at a given temperature. It is a fixed value when the temperature is constant and provides crucial information regarding the position of equilibrium. For the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \), the equation for Kc is similar to that of Q, but using concentrations at equilibrium: \[ K_{c} = \frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3} \].

When given the value of Kc, we know where the equilibrium lies. A large Kc value indicates a predominance of products; a small Kc value suggests a higher concentration of reactants at equilibrium. In the exercise, we're told that \( K_{c} = 62 \) which implies that at equilibrium, for this temperature, the concentration of ammonia will be quite high compared to nitrogen and hydrogen.

Molar Concentrations in Chemical Equilibrium

The molar concentration of a substance is defined as the number of moles of that substance divided by the volume of the solution or gas it is in, usually expressed in moles per liter (mol/L). Molar concentrations are central to calculating both Q and Kc because they quantify how much of each reactant and product is in the reaction mixture.

In the context of the given exercise, the concentrations provided are in millimoles per liter (mmol/L), a direct measure of the molarity of the species involved. Conversion from millimoles to moles (if required) is straightforward since 1 mmol is 0.001 mol. Having exact molar ratios allows chemists to precisely determine Q and predict whether a reaction will shift to form more products or reactants in pursuit of equilibrium. Regularly monitoring these concentrations can be vital to maintaining desired reaction conditions in both educational and industrial chemical processes.