The enthalpy and entropy change for the reaction- Br2 l - Cl2g - 2BrClg are 7-1 kcal mol and 25 calK-1mol-1 respectively- This reaction will be spontaneous at-

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Standard XII

Chemistry

Second Law of Thermodynamics

The enthalpy ...

Question

The enthalpy and entropy change for the reaction, $B r_{2} (l) + C l_{2} (g) \to 2 B r C l (g)$ are $7.1 k c a l$ $m o l$ and $25 c a l K^{- 1} m o l^{- 1}$ respectively. This reaction will be spontaneous at:

10^{o} C

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280 K

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290 K

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- 5^{o} C

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Solution

The correct option is C $290 K$
Using the relation, $Δ G = Δ H - T Δ S$
For a reaction to be spontaneous, $Δ G < 0$
$\Rightarrow Δ G = Δ H - T Δ S < 0$
$\Rightarrow 7100 - 25 T < 0$
$\Rightarrow T > 284 K$
Hence C is correct

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Similar questions

Q. The enthalpy and entropy change for the reaction:

B r_{2} (l) + C l_{2} (g) \to

B r C l (g)

are 30 kJ mol

^{- 1}

and 105 JK

^{- 1}

mol

^{- 1}

respectively. The temperature at which the reaction will be in equilibrium is:

Q. Calculate the free energy change at

298 K

for the reaction :

B r_{2} (l) + C l_{2} (g) \to 2 B r C l (g)

. For the reaction

Δ H^{o} = 29.3 k J

& the entropies of

B r_{2} (l), C l_{2} (g)

& BrCl(g) at the

298 K

are

152.3, 223.0, 239.7

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

respectively

Δ H

and

Δ S

for the reaction

{B r}_{2} (l) + {C l}_{2} (l) ⟶ 2 B r C l (g)

are

29.37 k J

and

104.0 J K^{- 1}

respectively. Above that temperature will this reaction become spontaneous?

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?

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

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The enthalpy and entropy change for the reaction, Br2(l)+Cl2(g)→2BrCl(g) are 7.1kcal mol and 25calK−1mol−1 respectively. This reaction will be spontaneous at:

The correct option is C 290KUsing the relation, ΔG=ΔH−TΔSFor a reaction to be spontaneous, ΔG<0⇒ΔG=ΔH−TΔS<0⇒7100−25T<0⇒T>284 KHence C is correct

The enthalpy and entropy change for the reaction, $B r_{2} (l) + C l_{2} (g) \to 2 B r C l (g)$ are $7.1 k c a l$ $m o l$ and $25 c a l K^{- 1} m o l^{- 1}$ respectively. This reaction will be spontaneous at:

The correct option is C $290 K$
Using the relation, $Δ G = Δ H - T Δ S$
For a reaction to be spontaneous, $Δ G < 0$
$\Rightarrow Δ G = Δ H - T Δ S < 0$
$\Rightarrow 7100 - 25 T < 0$
$\Rightarrow T > 284 K$
Hence C is correct