Sure! Here's a detailed, step-by-step solution to finding the standard free energy change ([tex]$\Delta G^{\circ}$[/tex]) for the reaction at [tex]\( 458^{\circ}C \)[/tex].
1. Initial Setup:
- We start with 1.00 mol of [tex]$H_2$[/tex] and 1.00 mol of [tex]$I_2$[/tex] in a 1.00 L flask.
- At equilibrium, the flask contains 0.225 mol of [tex]$H_2$[/tex], 0.225 mol of [tex]$I_2$[/tex], and 1.55 mol of [tex]$HI$[/tex].
2. Partial Pressures:
- Assuming the volume of the flask is 1.00 L, the partial pressures in the flask are proportional to the molar amounts.
- Thus, the partial pressures are:
- [tex]$P_{H_2} = 0.225$[/tex] atm
- [tex]$P_{I_2} = 0.225$[/tex] atm
- [tex]$P_{HI} = 1.55$[/tex] atm
3. Equilibrium Constant ([tex]$K_p$[/tex]):
- For the reaction: [tex]\( H_2 + I_2 \rightarrow 2 HI \)[/tex], the equilibrium constant in terms of partial pressures is given by:
[tex]\[
K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}}
\][/tex]
- Plugging in the values, we get:
[tex]\[
K_p = \frac{(1.55)^2}{0.225 \cdot 0.225}
\][/tex]
[tex]\[
K_p = \frac{2.4025}{0.050625}
\][/tex]
[tex]\[
K_p \approx 47.45679012345679
\][/tex]
4. Determining the Standard Free-Energy Change ([tex]$\Delta G^{\circ}$[/tex]):
- The standard free-energy change can be calculated using the relationship:
[tex]\[
\Delta G^{\circ} = -RT \ln(K_p)
\][/tex]
- Here, [tex]$R$[/tex] is the gas constant, [tex]\( 8.314 \; J/(mol \cdot K) \)[/tex], and [tex]$T$[/tex] is the temperature in Kelvin.
- The temperature in Kelvin is [tex]\( 458^{\circ}C + 273.15 = 731.15 \, K \)[/tex].
5. Calculating [tex]$\Delta G^{\circ}$[/tex]:
- First, we calculate the natural logarithm of [tex]$K_p$[/tex]:
[tex]\[
\ln(47.45679012345679) \approx 3.857
\][/tex]
- Now, we compute [tex]$\Delta G^{\circ}$[/tex]:
[tex]\[
\Delta G^{\circ} = -8.314 \, J/(mol \cdot K) \times 731.15 \, K \times 3.857
\][/tex]
[tex]\[
\Delta G^{\circ} \approx -23,462.99852761065 \, J/mol
\][/tex]
- Converting this to [tex]$kJ/mol$[/tex] by dividing by 1000:
[tex]\[
\Delta G^{\circ} \approx -23.463 \, kJ/mol
\][/tex]
Therefore, the standard free-energy change for the reaction at [tex]\( 458^{\circ}C \)[/tex] is:
[tex]\[
\Delta G^{\circ} \approx -23.463 \, kJ/mol
\][/tex]
