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Standard XII

Chemistry

Types of Equillibrium

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 30 kJ mol $^{- 1}$ and 105 JK $^{- 1}$ mol $^{- 1}$ respectively. The temperature at which the reaction will be in equilibrium is:

285.7K

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273 K

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450 K

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300 K

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Solution

The correct option is A 285.7K
$B r_{2} (l) + C l_{2} (g) ⟶ 2 B r C l (g)$

$△ H = 30 K J / m o l = 30 \times 10^{3} J / m o l$

$△ S = 105 J / K / m o l$

Since, $△ G = 0$ , then only system is at equilibrium.

So, $△ G = △ H - T △ S$

At equilibrium, $△ H = T △ S$

$T_{e q} = \frac{△ H}{△ S} = \frac{30 \times 10^{3}}{105} = 285.7 K$

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The enthalpy and entropy change for the reaction:Br2(l)+Cl2(g)→ 2BrCl(g)are 30 kJ mol−1 and 105 JK−1 mol−1 respectively. The temperature at which the reaction will be in equilibrium is:

The correct option is A 285.7KBr2(l)+Cl2(g)⟶2BrCl(g)△H=30KJ/mol=30×103J/mol△S=105J/K/molSince, △G=0, then only system is at equilibrium.So, △G=△H−T△SAt equilibrium, △H=T△STeq=△H△S=30×103105=285.7K

The enthalpy and entropy change for the reaction:
$B r_{2} (l) + C l_{2} (g) \to$ 2 $B r C l (g)$
are 30 kJ mol $^{- 1}$ and 105 JK $^{- 1}$ mol $^{- 1}$ respectively. The temperature at which the reaction will be in equilibrium is: