To solve for [tex]\(\Delta H\)[/tex] at [tex]\(T = 290\, \text{K}\)[/tex], we use the relationship between enthalpy change ([tex]\(\Delta H\)[/tex]), entropy change ([tex]\(\Delta S\)[/tex]), and temperature ([tex]\(T\)[/tex]) in thermodynamics. The expression that ties these quantities together is:
[tex]\[
\Delta H(T) = \Delta H - T \Delta S
\][/tex]
Given values:
- [tex]\(\Delta H = -120\, \text{kJ}\)[/tex]
- [tex]\(\Delta S = -150\, \text{J/K}\)[/tex]
- [tex]\(T = 290\, \text{K}\)[/tex]
#### Step-by-Step Solution:
1. Convert [tex]\(\Delta S\)[/tex] to the same units as [tex]\(\Delta H\)[/tex]:
Since [tex]\(\Delta H\)[/tex] is given in kJ and [tex]\(\Delta S\)[/tex] in J/K, we need to convert [tex]\(\Delta S\)[/tex] from J/K to kJ/K. We do this by dividing [tex]\(\Delta S\)[/tex] by 1000 (since [tex]\(1\, \text{kJ} = 1000\, \text{J}\)[/tex]):
[tex]\[
\Delta S_{\text{kJ/K}} = \frac{\Delta S}{1000} = \frac{-150\, \text{J/K}}{1000} = -0.15\, \text{kJ/K}
\][/tex]
2. Substitute the given values into the formula:
[tex]\[
\Delta H(T) = \Delta H - T \Delta S_{\text{kJ/K}}
\][/tex]
Given [tex]\(\Delta H = -120\, \text{kJ}\)[/tex], [tex]\(T = 290\, \text{K}\)[/tex], and [tex]\(\Delta S_{\text{kJ/K}} = -0.15\, \text{kJ/K}\)[/tex]:
[tex]\[
\Delta H(290\, \text{K}) = -120\, \text{kJ} - 290\, \text{K} \times (-0.15\, \text{kJ/K})
\][/tex]
3. Calculate the term involving [tex]\(T\)[/tex] and [tex]\(\Delta S\)[/tex]:
[tex]\[
290\, \text{K} \times (-0.15\, \text{kJ/K}) = -43.5\, \text{kJ}
\][/tex]
4. Combine the terms to find [tex]\(\Delta H(290\, \text{K})\)[/tex]:
[tex]\[
\Delta H(290\, \text{K}) = -120\, \text{kJ} - (-43.5\, \text{kJ})
\][/tex]
[tex]\[
\Delta H(290\, \text{K}) = -120\, \text{kJ} + 43.5\, \text{kJ}
\][/tex]
5. Simplify the expression:
[tex]\[
\Delta H(290\, \text{K}) = -76.5\, \text{kJ}
\][/tex]
So, the change in enthalpy [tex]\(\Delta H\)[/tex] at [tex]\(290\, \text{K}\)[/tex] for the given reaction is [tex]\(-76.5\, \text{kJ}\)[/tex].
