Question

enthalpy and entropy changes of a reaction are 40.63 KJ /mol and 108.8Jk^-1Mol^-1 respectively. calculate the value of delta ∆G.???​

Answer 1

Answer:

✓change in enthalpy (∆H) = 40.63kJ/Mol

✓change in entropy (∆S) = 108.8 J k^-1 Mol^-1

✓Temperature = 273 + 27 = 300 k

So...

∆G = ∆H - T∆S

= 40.63 - 300 x 108.8 / 1000

= 40.63 - 3 x 10.88

= 40.63 - 32.64

= 7.99kJ

So...∆G = 7.99kJ

Answer 2

Explanation:

Heya mate

change in enthalpy= 40.63kj/mol (given )

change in entropy= 108.8jk^-1. (given)

we have to find∆G ,

Temperature at Kelvin= 273+30= 300k

∆G=∆H-T∆S

= 40.63-300*108.8

= 7.99kj