For a reaction at 25 -C enthalpy change - H and entropy change - S are -11-7- 103 J mol-1 and -105 J mol-1 K -1 respectively- The reaction is-

As we know, Δ G =Δ H TΔ S ∴Δ G = 11700 298× ( 105)=+19590 J =19.59 kJ Thus, reaction is non spontaneous. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

For a reactio...

Question

For a reaction at $25^{\circ} C$ enthalpy change $(Δ H)$ and entropy change $(Δ S)$ are $- 11.7 \times 10^{3} {J mol}^{- 1}$ and $- 105 {J mol}^{- 1} K^{- 1}$ respectively. The reaction is:

Spontaneous

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

Non-spontaneous

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

Instantaneous

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

None of these

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

Open in App

Solution

The correct option is B Non-spontaneous
As we know,
$Δ G = Δ H - T Δ S$
$∴ Δ G = - 11700 - 298 \times (- 105) = + 19590 J = 19.59 kJ$
Thus, reaction is non-spontaneous.

Suggest Corrections

Similar questions

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:

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:

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 K J $m o l^{- 1}$ and -105 J $m o l^{- 1}$ $K^{- 1}$ , respectively. It is a spontaneous reaction

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?

Join BYJU'S Learning Program

Explore more

Derivation of Kp and Kc

Standard XII Chemistry

Join BYJU'S Learning Program

For a reaction at 25 ∘C enthalpy change (ΔH) and entropy change (ΔS) are −11.7×103 J mol−1 and −105 J mol−1 K−1 respectively. The reaction is:

The correct option is B Non-spontaneousAs we know, ΔG=ΔH−TΔS ∴ΔG=−11700−298×(−105)=+19590 J=19.59 kJ Thus, reaction is non-spontaneous.

For a reaction at $25^{\circ} C$ enthalpy change $(Δ H)$ and entropy change $(Δ S)$ are $- 11.7 \times 10^{3} {J mol}^{- 1}$ and $- 105 {J mol}^{- 1} K^{- 1}$ respectively. The reaction is:

The correct option is B Non-spontaneous
As we know,
$Δ G = Δ H - T Δ S$
$∴ Δ G = - 11700 - 298 \times (- 105) = + 19590 J = 19.59 kJ$
Thus, reaction is non-spontaneous.