\(\frac{1}{2}\)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?
88.71 KJ
85.48 KJ
-204.00 KJ
-3427.40 KJ
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The change in free energy (∆G°) for a reaction can be calculated using the formula: ∆G° = ∆H° - T∆S°. In this case, ∆H° is given as 89 kJ/mol and ∆S° is given as 11.8 J/mol/K. The temperature (T) is given as 25°C, which is equivalent to 298.15 K. Plugging these values into the formula gives us:
∆G° = 89 kJ/mol - (298.15 K)(11.8 J/mol/K)(1 kJ/1000 J) ∆G° = 89 kJ/mol - (3.52 kJ/mol) ∆G° = 85.48 kJ/mol
So the correct answer is B. 85.48 KJ.
^G = ^H - T^S. 89 - (298*11.8)/1000. Because we have to convert the unit of ^S from joules (J) to kilojoules (kJ). So the ans is B.
y dis myschool, y all dis errors in some of ur ansas, dis should be corrected pls
Apologies please! Thank you all for your contributions.
Kindly note that the necessary correction has been effected.
∆G = ∆H - T(∆S)
∆G = 89 - 24 + 273(11.8)
∆G = 89 - 298 * 11.8
∆G = 89 - 3516.4
∆G = -3,427.4
kindly go through the formula and calculation I hope this will help
B is correct pls myschool make corrections
the enthalpy is to be converted from Kj to J
