The equilibrium constant for the reaction - Br 2 - 2Br at 500 K and 700 K are 1 - 10 -10 and 1 - 10 -5 respectively- The reaction is-

The correct option is A endothermic Br_2⇌ 2Br K_d=[Br]^2[Br_2] (i) at 500K⇒ K_d=10^ 10 (ii) at 700K⇒ K_d=10^ 5 As temperature increases ...

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Byju's Answer

Standard XII

Chemistry

Auto Protolysis of Water

The equilibri...

Question

The equilibrium constant for the reaction :

${B r}_{2}$ $⇌$ 2Br

at 500 K and 700 K are 1 $\times$ $10^{- 10}$ and 1 $\times$ $10^{- 5}$ respectively. The reaction is:

endothermic

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exothermic

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fast

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slow

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Solution

The correct option is A endothermic
$B r_{2} ⇌ 2 B r$

$K_{d} = \frac{[B r]^{2}}{[B r_{2}]}$

(i) at $500 K \Rightarrow K_{d} = 10^{- 10}$

(ii) at $700 K \Rightarrow K_{d} = 10^{- 5}$

As temperature increases $K_{d}$ also increases

$\Rightarrow [B r]$ also increases

$\Rightarrow$ The reaction is endothermic as increasing temperature
leads to more products formed.

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The equilibrium constant for the reaction : Br2 ⇌ 2Br at 500 K and 700 K are 1 × 10−10 and 1 × 10−5 respectively. The reaction is:

The correct option is A endothermicBr2⇌2BrKd=[Br]2[Br2](i) at 500K⇒Kd=10−10(ii) at 700K⇒Kd=10−5As temperature increases Kd also increases⇒[Br] also increases⇒ The reaction is endothermic as increasing temperature leads to more products formed.

The equilibrium constant for the reaction :

${B r}_{2}$ $⇌$ 2Br

at 500 K and 700 K are 1 $\times$ $10^{- 10}$ and 1 $\times$ $10^{- 5}$ respectively. The reaction is: