Question

If the reaction \(\mathrm{Fe}_{2} \mathrm{N}(s)+3 / 2 \mathrm{H}_{2}(g) \rightleftharpoons\) \(2 \mathrm{Fe}(s)+\mathrm{NH}_{3}(g)\) comes to equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700 . and 800 . K. \(P_{N H} / P_{H_{2}}=2.165\) and \(1.083,\) respectively, if only \(\mathrm{H}_{2}(g)\) was initially present in the gas phase and \(\mathrm{Fe}_{2} \mathrm{N}(s)\) was in excess. a. Calculate \(K_{P}\) at \(700 .\) and \(800 .\) K. b. Calculate \(\Delta S_{R}^{\circ}\) at \(700 . \mathrm{K}\) and \(800 . \mathrm{K}\) and \(\Delta H_{R}^{\circ}\) assuming that it is independent of temperature. c. Calculate \(\Delta G_{R}^{\circ}\) for this reaction at \(298.15 \mathrm{K}\)

Short answer

In summary, the equilibrium constants at 700 K and 800 K are \(K_P = 8.566\) and \(K_P = 1.186\), respectively. The standard enthalpy change for the reaction is \(\Delta H^{\circ}_R = 20,720 ~\text{J/mol}\), and the standard entropy changes at 700 K and 800 K are \(\Delta S^{\circ}_R = 27.61 ~\text{J/mol K}\) and \(\Delta S^{\circ}_R = 25.98 ~\text{J/mol K}\), respectively. The standard Gibbs free energy change for the reaction at \(298.15\ \mathrm{K}\) is \(\Delta G^{\circ}_{\mathrm{R}} = 11,470\ \mathrm{J/mol}\).

Step-by-step solution

Write the reaction and the expression for \(K_P\)

The reaction is given by: \[\mathrm{Fe}_{2} \mathrm{N}(s)+\frac{3}{2}\mathrm{H}_{2}(g)\rightleftharpoons 2 \mathrm{Fe}(s)+\mathrm{NH}_{3}(g)\] The expression for the equilibrium constant in terms of partial pressures, \(K_P\), is: \[K_P = \frac{\left(\frac{P_\mathrm{NH_3}}{P^\circ}\right)}{\left(\frac{P_\mathrm{H_2}}{P^\circ}\right)^{\frac{3}{2}}}\] where \(P_{NH_3}\) and \(P_{H_2}\) are the partial pressures of ammonia and hydrogen at equilibrium, and we are using standard pressure \(P^\circ = 1\) bar.

Calculate the equilibrium pressure of ammonia and hydrogen

We are given that the ratio of partial pressures of ammonia and hydrogen at 700 K and 800 K is \(2.165\) and \(1.083\) respectively: \[\frac{P_{NH_3}}{P_{H_2}} = 2.165\) at 700 K, \[\frac{P_{NH_3}}{P_{H_2}} = 1.083\) at 800 K. We also know that the total pressure at equilibrium is 1 bar, which means: \[P_{NH_3} + P_{H_2} = 1\] We can solve these equations simultaneously to obtain the values of \(P_{NH_3}\) and \(P_{H_2}\) at each temperature: At 700 K: \[P_{NH_3} = 2.165 P_{H_2}\] \[P_{NH_3} + P_{H_2} = 1\] \[2.165 P_{H_2} + P_{H_2} = 1\] \[3.165 P_{H_2} = 1\] \[P_{H_2} = \frac{1}{3.165} = 0.316\] \[P_{NH_3} = 2.165 \times 0.316 = 0.684\] At 800 K: \[P_{NH_3} = 1.083 P_{H_2}\] \[P_{NH_3} + P_{H_2} = 1\] \[1.083 P_{H_2} + P_{H_2} = 1\] \[2.083 P_{H_2} = 1\] \[P_{H_2} = \frac{1}{2.083} = 0.480\] \[P_{NH_3} = 1.083 \times 0.480 = 0.520\] We now have the partial pressures of ammonia and hydrogen at each temperature.

Calculate \(K_P\) at 700 K and 800 K

Using the values we just found, we will plug into the equation for \(K_P\): At 700 K: \[K_P = \frac{(0.684)}{(0.316)^{\frac{3}{2}}} = 8.566\] At 800 K: \[K_P = \frac{(0.520)}{(0.480)^{\frac{3}{2}}} = 1.186\] Therefore, the equilibrium constants at 700 K and 800 K are \(K_P = 8.566\) and \(K_P = 1.186\), respectively. #b. Calculate \(\Delta S^{\circ}_R\) at \(700K\) and \(800K\) and \(\Delta H^{\circ}_R\) assuming that it is independent of temperature#

Use the van't Hoff equation

The van't Hoff equation relates the change in the equilibrium constant with temperature: \[\frac{\mathrm{d}\ln{K}}{\mathrm{d}T} = \frac{\Delta H^{\circ}_R}{RT^2}\] Since \(\Delta H^{\circ}_R\) is assumed to be independent of temperature, we can integrate the van't Hoff equation between 700 K and 800 K: \[\int_{K_1}^{K_2} \frac{\mathrm{d}(\ln{K})}{\mathrm{d}T} \mathrm{d}T = \int_{T_1}^{T_2} \frac{\Delta H^{\circ}_R}{RT^2}\mathrm{d}T\] \[\ln{\frac{K_2}{K_1}} = -\frac{\Delta H^{\circ}_R}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] We can now plug in the values for the equilibrium constants and temperatures to determine \(\Delta H^{\circ}_R\).

Calculate \(\Delta H^{\circ}_R\)

Plugging the values: \[\ln{\frac{1.186}{8.566}} = -\frac{\Delta H^{\circ}_R}{8.314}\left(\frac{1}{700} - \frac{1}{800}\right)\] We can then solve for \(\Delta H^{\circ}_R\): \[\Delta H^{\circ}_R = -(9.035 \times 10^3)(-2.292) = 20,720 ~\text{J/mol}\]

Calculate \(\Delta S^{\circ}_R\) at \(700K\) and \(800K\)

Using the relationship between the standard Gibbs free energy change, standard enthalpy change, and standard entropy change: \[\Delta G^{\circ}_R = \Delta H^{\circ}_R - T\Delta S^{\circ}_R\] At equilibrium, \(\Delta G^{\circ}_R = -RT\ln{K_P}\), so: \[-RT\ln{K_P} = \Delta H^{\circ}_R - T\Delta S^{\circ}_R\] We can solve for \(\Delta S^{\circ}_R\) at 700 K and 800 K: At 700 K: \[\Delta S^{\circ}_R = \frac{\Delta H^{\circ}_R + RT\ln{K_P}}{T} = \frac{20,720 + (700)(8.314)\ln{8.566}}{700} = 27.61 ~\text{J/mol K}\] At 800 K: \[\Delta S^{\circ}_R = \frac{\Delta H^{\circ}_R + RT\ln{K_P}}{T} = \frac{20,720 + (800)(8.314)\ln{1.186}}{800} = 25.98 ~\text{J/mol K}\] #c. Calculate \(\Delta G^{\circ}_R\) for this reaction at \(298.15K\)#

Extrapolate \(\Delta S^{\circ}_R\) to \(298.15K\)

We will assume the change in entropy is linear with temperature: \[\Delta S^{\circ}_R(298.15) = \Delta S^{\circ}_R(700) - \frac{\Delta S^{\circ}_R(800) - \Delta S^{\circ}_R(700)}{800-700}(700-298.15)\] \[\Delta S^{\circ}_R(298.15) = 27.61 - \frac{25.98 - 27.61}{100}(401.85) = 29.27 ~\text{J/mol K}\]

Calculate \(\Delta G^{\circ}_R\) at \(298.15K\)

Now calculate the standard Gibbs free energy change at this temperature: \[\Delta G^{\circ}_R(298.15) = \Delta H^{\circ}_R - T\Delta S^{\circ}_R = 20,720 - 298.15(29.27) = 11,470 ~\text{J/mol}\] Thus, \(\Delta G^{\circ}_{\mathrm{R}}\) for this reaction at \(298.15\ \mathrm{K}\) is \(11,470\ \mathrm{J/mol}\).

Key concepts

Equilibrium Constant (Kp)

In the realm of chemical reactions, an equilibrium constant denoted by the symbol K_p is fundamentally important. This constant is a measure of the extent to which a chemical reaction proceeds to reach a state of equilibrium at a given temperature. In gas-phase reactions, where the substances involved are under varying pressures, K_p is particularly useful as it is defined in terms of partial pressures of the gaseous reactants and products.

The expression for K_p in a reaction involving gases is derived from the balanced chemical equation and takes the form of a ratio of the products' partial pressures raised to the power of their coefficients to the reactants' partial pressures also raised to the power of their respective coefficients. This ratio describes how the product's concentration compares to the reactants' when the reaction has reached its equilibrium state. Knowing the value of K_p can significantly elucidate the reaction's behavior under various conditions, including temperature changes.

Partial Pressure

The concept of partial pressure is integral to understanding gas-phase chemical reactions at equilibrium. It refers to the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume of the mixture at the same temperature.

Within a chemical reaction, each gas' partial pressure is directly proportional to its mole fraction in the mixture, making it a crucial component in calculating the equilibrium constant K_p. In practical terms, it allows us to determine how the presence of different gases influences the position of equilibrium. The partial pressures are commonly used in calculations involving the equilibrium constant because they reflect the concentrations of the gases present, which vary according to the reaction's progression towards equilibrium.

Van't Hoff Equation

When studying how equilibrium shifts with temperature, the van't Hoff equation proves particularly enlightening. It establishes the relationship between the changes in temperature and the corresponding changes in the equilibrium constant.

The equation posits that the natural logarithm of the equilibrium constant K_p changes linearly with the inverse of the temperature (in Kelvin). van't Hoff provides a quantitative method to predict the effect of temperature change on reaction rates and equilibria. This principle is central to understanding chemical kinetics and thermodynamics, especially when it comes to predicting the direction in which a reaction's equilibrium will shift when there are changes in temperature.

Gibbs Free Energy

Among the pillars of thermodynamics is the Gibbs free energy, denoted by G. It offers a window into the spontaneity of a reaction under constant pressure and temperature conditions. In essence, Gibbs free energy is a thermodynamic potential that can predict the direction of chemical processes and the feasibly achievable work.

The change in Gibbs free energy, represented as ΔG, during a reaction helps determine whether a reaction is thermodynamically favorable or not. If ΔG is negative, the reaction is spontaneous, while a positive value indicates that it is non-spontaneous. When ΔG equals zero, the system is in equilibrium, and no net reaction occurs. For reactions at equilibrium, ΔG also relates directly to the equilibrium constant K_p, underlining its integral role in chemical thermodynamics.