\(\frac{1}{2}\)N\(_2\)(g) + \(\frac{1}{2}\)O\(_2\)(g) → NO(g) ∆H° = 89 KJ mol\(^{-1}\). If the entropy change for the reaction above at 25°C is 11.8 J mol\(^{-1}\), calculate the change in free energy. ∆G°, for the reaction at 25°C?
\(\frac{1}{2}\)N\(_2\)(g) + \(\frac{1}{2}\)O\(_2\)(g) → NO(g) ∆H° = 89 KJ mol\(^{-1}\)
First, ensure the right conversion is made.
T\(^0\)C = 25 + 273 = 298K
∆S = 11.8 = \(\frac{11.8}{1000}\) = 0.0118 kJmol\(^{-1}\) (Conversion from J/mol to kJ/mol)
∆H = 89 kJmol\(^{-1}\)
Applying the formula; ∆G\(^0\) = ∆H - T∆S
= 89 - (298 x 0.0118)
= 89 - 3.5164
= 85.4836
= 85.48 kJmol\(^{-1}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}