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T (in ,K) {1}{T} (in , K^{-1}) _{10} K 7691.3 10^{-3} 2.9 667

Question

$T (in K)$ $\frac{1}{T} (in K^{- 1})$ $lo g_{10} K$
$769$ $1.3 \times 1 0^{- 3}$ $2.9$
$667$ $1.5 \times 1 0^{- 3}$ $1.1$
From the following data: the activation energy for the reaction (cal/mol) $H_{2} + I_{2} \to 2 H I$

4 \times 10^{4}

2 \times 10^{4}

8 \times 10^{4}

3 \times 10^{4}

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Chloe Discussed

From the following data: the activation energy for the reaction (cal/mol)

H_{2} + I_{2} \to 2 H I

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9 mins ago

Text solutionVerified

lo g \frac{K _{2}}{K _{1}} = \frac{E _{a c t}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

\Rightarrow 2.9 - 1.1 = \frac{E _{a c t}}{2} \times 0.2 \times 1 0^{- 3}

\Rightarrow E_{a c t} = \frac{2 \times 1.8 \times 1 0 ^{3} \times 10}{2}

= 18000 c a l

\approx 2 \times 1 0^{4} c a l

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Question Text	From the following data: the activation energy for the reaction (cal/mol) $H_{2} + I_{2} \to 2 H I$
Answer Type	Text solution:1
Upvotes	150