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The rate of a reaction gets doubled when the temperature changes from $7^{\circ}C$ to $17^{\circ}$ . By what factor will it change for the temperature change from $17^{\circ}C$ to $27^{\circ}C$

$(A)\;1.81 \\ (B)\;1.71 \\(C)\;1.91 \\(D)\;1.76 $
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$2.303 \log \large\frac{K_2}{K_1}=\frac{E_A}{R} \bigg[\large\frac{T_2-T_1}{T_1T_2}\bigg]$
$T= 280 \;K; T= 290\;K; \large\frac{K_2}{K_1} $$=2$
Find $E_a$ and then again use,
$2.303 \log \large\frac{K_2'}{K_1'}=\frac{E_A}{R} \bigg[\large\frac{T_2'-T_1'}{T_1'T_2'}\bigg]$
If $ T_1'=290\;K;T_2'= 300\;k$
Hence c is the correct answer.
answered Apr 3, 2014 by meena.p

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