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Question

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 $$27^{0}C$$?


A
1.81
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B
1.71
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C
1.91
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D
1.76
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Solution

The correct option is C 1.91
$$log_{ 10 }\displaystyle \frac { k_{ 2 } }{ k_{ 1 } } =\displaystyle \frac { 2.303E_a }{ R } \left( \displaystyle \frac { T_{ 2 }-T_{ 1 } }{ T_{ 1 }T_{ 2 } }  \right) \\ \displaystyle \frac { R\ log\ 2\times 280\times 290 }{ 2.303\times 10 } = E_a\\ log \displaystyle \frac { k_{ 4 } }{ k_{ 3 } } =\displaystyle \frac { 2.303\times R\ log\ 2\times 280\times 290 }{ 2.303\times 10\times R } \left( \displaystyle \frac { 10 }{ 290 \times 300 }  \right) \\ log_{ 10 }\left(\displaystyle \frac { k_{ 4 } }{ k_{ 3 } }  \right) =\displaystyle \frac { 280 }{ 300 } log\ 2\quad\\ \displaystyle \frac { k_{ 4 } }{ k_{ 3 }}=1.91$$

Chemistry

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