The rate of a reaction gets doubled when the temperature changes from $7^{0} C$ to $17^{0} C$ . By which factor will it change for the temperature change from $17^{0} C$ to $27^{0} C$ ?

1.81

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1.71

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1.91

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1.76

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Solution

The correct option is C 1.91
$l o g_{10} \frac{k_{2}}{k_{1}} = \frac{2.303 E_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \frac{R l o g 2 \times 280 \times 290}{2.303 \times 10} = E_{a} l o g \frac{k_{4}}{k_{3}} = \frac{2.303 \times R l o g 2 \times 280 \times 290}{2.303 \times 10 \times R} (\frac{10}{290 \times 300}) l o g_{10} (\frac{k_{4}}{k_{3}}) = \frac{280}{300} l o g 2 \frac{k_{4}}{k_{3}} = 1.91$

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The rate of a reaction gets doubled when the temperature changes from 70C to 170C. By which factor will it change for the temperature change from 170C to 270C?

The correct option is C 1.91log10k2k1=2.303EaR(T2−T1T1T2)R log 2×280×2902.303×10=Ealogk4k3=2.303×R log 2×280×2902.303×10×R(10290×300)log10(k4k3)=280300log 2k4k3=1.91

The rate of a reaction gets doubled when the temperature changes from $7^{0} C$ to $17^{0} C$ . By which factor will it change for the temperature change from $17^{0} C$ to $27^{0} C$ ?