Question

standard enthalpy and standard entropy changes for the oxidation of ammonia at 298k are 382.64kj /mole and 145.6jmol respectively standard gibbs energy change for same reaction at 298k

Answer 1

Given conditions ⇒

Standard Enthalpy(H) = 382640 J/mole.

Standard Entropy(S) = 145.6 J/mole.

Temperature(T) = 298 K.

Using the Formula,

 G = H - TS

⇒ G = 382640 - 145.6 × 298

⇒ G = 382640 - 43388.8

∴ G = 339.251 kJ/mole.

Hope it helps .

Answer 2

Answer:

Given conditions ⇒

Standard Enthalpy(H) = -382640 J/mole.

Standard Entropy(S) = 145.6 J/mole.

Temperature(T) = 298 K.

Using the Formula,  

G = H - TS ⇒ G = 382640 - 145.6 × 298⇒ G = 382640 - 43388.8

∴ G = -339.251 kJ/mole.