Question 23.5: The reaction system described by the chemical equation 2HI(g......
The reaction system described by the chemical equation
2\ HI(g) ⇋ H_{2}(g) + I_{2}(g)is prepared at 598 K with the initial pressures P_{HI} = 0.500\ bar, P_{H_{2}} = 0.150\ bar, and P_{I_{2}}= 4.25 × 10^{–2}\ bar. Given that K_{p} = 1.08 × 10^{–2}\ at\ 598\ K, calculate the value of ΔG_{rxn} for this reaction system and indicate in which direction the reaction as written will proceed spontaneously.
The value of Q_{p} for this equation is given by
Q_{p}=\frac{P_{H_{2}}P_{I_{1}}}{(P_{HI})^2} =\frac{(0.150\ bar)(4.25 × 10^{–2}\ bar)}{(0.500\ bar)^2} = 0.0255Thus,
∆G_{rxn} = RT\ ln\left(\frac{Q_{p}}{K_{p}} \right) =(8.3145\ J·K^{–1}·mol^{–1})(598\ K)ln\left(\frac{0.0255}{1.08 × 10^{–2}} \right) \\ =4.27\ kJ·mol^{–1}where, like the other thermodynamic variables we encountered in Chapter 14, the subscript rxn and the unit of mol^{–1} here refer to one mole of the chemical equation as written. Also recall from Chapter 19 that the values of Q_{p} and K_{p} (or Q_{c} and K_{c}) depend upon how we choose to write the chemical equation (see Problem 23-62).
The positive value of ∆G_{rxn} (which results from the fact that Q_{p} > K_{p}) implies that the reaction system will evolve in such a way that the concentrations of H_{2}(g) and I_{2}(g) will decrease and that of HI(g) will increase. That is, the reaction described by
2\ HI(g,\ 0.500\ bar) ⇋ H_{2}(g,\ 0.150\ bar) + I_{2}(g,\ 0.0425\ bar)proceeds spontaneously from right to left as written.