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Q.

The equilibrium constants for the reaction, ${Br}_{2} ⇌ 2 Br$ at 500 K and 700 K are 1x10^-10 and 1 x 10^-5 respectively. The reaction is:

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a

endothermic

b

exothermic

c

fast

d

slow

answer is A.

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Detailed Solution

$2 .303 \log \frac{K_{2}}{K_{1}} = \frac{ΔH}{R} = [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$∴$ K increases with temperature and thus, $ΔH = + ve$

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The equilibrium constants for the reaction, Br2⇌2Br at 500 K and 700 K are 1x10-10 and 1 x 10-5 respectively. The reaction is: