For the melting of ice at $25^{\circ} C$ , the enthalpy of fusion is $6.97 K J m o l^{- 1}$ and entropy of fusion is $25.4 J K^{- 1} g^{- 1}$ . The free energy change and nature of the reaction at this temperature is :

– 599.2 J

, spontaneous

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 599.2 J

, non spontaneous

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14.54 K J

, non spontaneous

No worries! We‘ve got your back. Try BYJU‘S free classes today!

– 14.54 KJ

, spontaneous

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A – 599.2 J , spontaneous
$Δ H_{f u s i o n} = 6.97 k J m o l^{- 1} = 6970 J m o l^{- 1} Δ H_{f u s i o n} = 25.4 J K^{- 1} m o l^{- 1}, T_{f} = 25 + 273 = 398 K According to Gibbs Helmholtz equation, Δ G = Δ H - T Δ S = (6970 J m o l^{- 1}) - (298 K) \times (25.4 J K^{- 1} m o l^{- 1}) = 6970 - 7569.2 = - 599.2 J m o l^{- 1}$

Suggest Corrections

Similar questions

Q. Zinc reacts with dilute hydrochloric acid to give hydrogen gas at

17^{o} C .

The enthalpy change of the reaction is

- 12.55 k J m o l^{- 1}

and entropy change equals to

5.0 J K^{- 1} m o l^{- 1}

for the reaction. Calculate the free energy change and predict whether the reaction is spontaneous or not.

The $△ H$ and $△ S$ for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 k J $k^{- 1}$ respectively. The temperature at which the free energy change will be zero and below of this temperature the nature of reaction would be

Q. For a reaction at

25^{\circ} C

enthalpy change

(Δ H)

and entropy change

(Δ S)

are

- 11.7 \times 10^{3} J m o l^{- 1}

and

- 105 J m o l^{- 1} K^{- 1}

respectively. The reaction is:

For a reaction at $25^{\circ} C$ enthalpy change $(△ H)$ and entropy change $(△ S)$ are $- 11.7 \times 103 J m o l^{- 1}$ and $- 105 J m o l^{- 1} K^{- 1}$ respectively. Find out whether this reaction is spontaneous or not?

Q. Calculate the change in entropy for fusion of 1 mole of ice. The melting point of ice is 273 K and molar enthalpy of fusion for ice = 6.0 kJ

m o l e^{- 1}

Join BYJU'S Learning Program

For the melting of ice at 25∘C, the enthalpy of fusion is 6.97KJmol−1 and entropy of fusion is 25.4JK−1g−1. The free energy change and nature of the reaction at this temperature is :

The correct option is A – 599.2 J , spontaneousΔHfusion=6.97kJmol−1=6970Jmol−1ΔHfusion=25.4JK−1mol−1,Tf=25+273=398KAccording to Gibbs Helmholtz equation,ΔG=ΔH−TΔS=(6970Jmol−1)−(298K)×(25.4JK−1mol−1)=6970−7569.2=−599.2Jmol−1

For the melting of ice at $25^{\circ} C$ , the enthalpy of fusion is $6.97 K J m o l^{- 1}$ and entropy of fusion is $25.4 J K^{- 1} g^{- 1}$ . The free energy change and nature of the reaction at this temperature is :