Consider the reaction X₂(g) → 2X(g). When a vessel initially containing 755 torr of X2 comes to equilibrium at 298 K, the equilibrium partial pressure of X is 103 torr. The same reaction is repeated with an initial partial pressure of 748 torr of X₂ at 755 K; the equilibrium partial pressure of X is 532 torr. Find ΔH° for the reaction.

Name: Chemistry: A Molecular Approach
Author: Tro
ISBN: 9780134112831

Verified step by step guidance

insert step 1> Start by writing the expression for the equilibrium constant (Kp) for the reaction X2(g) → 2X(g). The expression is Kp = (P_X)^2 / P_X2, where P_X is the partial pressure of X and P_X2 is the partial pressure of X2.

insert step 2> For the first condition at 298 K, calculate the equilibrium partial pressure of X2 using the initial pressure and the change in pressure due to the formation of X. The change in pressure is half of the equilibrium pressure of X, so P_X2 = initial pressure - (1/2) * P_X.

insert step 3> Calculate Kp at 298 K using the equilibrium pressures found in step 2.

insert step 4> Repeat steps 2 and 3 for the second condition at 755 K to find the equilibrium constant Kp at this temperature.

insert step 5> Use the van 't Hoff equation, ln(K2/K1) = -ΔH°/R * (1/T2 - 1/T1), where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively, and R is the gas constant, to solve for ΔH°.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium and Partial Pressure

In a chemical reaction at equilibrium, the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products. Partial pressure refers to the pressure exerted by a single component in a mixture of gases. Understanding how to calculate and interpret partial pressures is crucial for analyzing equilibrium conditions in gas-phase reactions.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in equilibrium constant (K) with temperature to the enthalpy change (ΔH°) of a reaction. It is expressed as ln(K2/K1) = -ΔH°/R(1/T2 - 1/T1), where R is the gas constant. This equation is essential for determining the enthalpy change of a reaction by comparing equilibrium constants at different temperatures.

Enthalpy Change (ΔH°)

Enthalpy change (ΔH°) is the heat content change of a system at constant pressure during a chemical reaction. It indicates whether a reaction is exothermic (releases heat, ΔH° < 0) or endothermic (absorbs heat, ΔH° > 0). Calculating ΔH° is vital for understanding the energy dynamics of a reaction and predicting how temperature changes affect equilibrium.

Related Practice

Textbook Question

Indicate and explain the sign of ΔS_univ for each process. a. 2 H₂(g) + O₂(g) → 2 H₂O (l) at 298 K.

Dinitrogen tetroxide decomposes to nitrogen dioxide: N₂O₄(g) → 2 NO₂(g) ΔH°_rxn = 55.3 kJ At 298 K, a reaction vessel initially contains 0.100 atm of N₂O₄. When equilibrium is reached, 58% of the N₂O₄has decomposed to NO₂. What percentage of N₂O₄decomposes at 388 K? Assume that the initial pressure of N₂O₄is the same (0.100 atm).

All the oxides of nitrogen have positive values of ΔG°_f at 298 K, but only one common oxide of nitrogen has a positive ΔS°_f. Identify that oxide of nitrogen without reference to thermodynamic data and explain.

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Textbook Question

These reactions are important in catalytic converters in automobiles. Calculate ΔG° for each at 298 K. Predict the effect of increasing temperature on the magnitude of ΔG°.

a. 2 CO(g) + 2 NO(g) → N₂(g) + 2 CO₂(g)

b. 5 H₂(g) + 2 NO(g) → 2 NH₃(g) + 2 H₂O(g)

c. 2 H₂(g) + 2 NO(g) → N₂(g) + 2 H₂O(g)

d. 2 NH₃(g) + 2 O₂(g) → N₂O(g) + 3 H₂O(g)

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Textbook Question

The values of ΔG°_f for the hydrogen halides become less negative with increasing atomic number. The ΔG°_f of HI is slightly positive. However, the trend in ΔS°_f is to become more positive with increasing atomic number. Explain.

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