JEE MAIN - Chemistry (2005 - No. 40)
t1/4 can be taken as the time taken for the concentration of a reactant to drop to $$3 \over 4$$ of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be written as
0.10 / K
0.29 / K
0.69 / K
0.75 / K
Explanation
$${t_{1/4}} = {{2.303} \over K}\log {1 \over {3/4}}$$
$$ = {{2.303} \over K}\log {4 \over 3}$$
$$ = {{2.303} \over K}\left( {\log \,4 - \log 3} \right)$$
$$ = {{2.303} \over K}\left( {2{{\log }^2} - \log 3} \right)$$
$$ = {{2.303} \over K}\left( {2 \times 0.301 - 0.4771} \right)$$
$$ = {{0.29} \over K}$$
$$ = {{2.303} \over K}\log {4 \over 3}$$
$$ = {{2.303} \over K}\left( {\log \,4 - \log 3} \right)$$
$$ = {{2.303} \over K}\left( {2{{\log }^2} - \log 3} \right)$$
$$ = {{2.303} \over K}\left( {2 \times 0.301 - 0.4771} \right)$$
$$ = {{0.29} \over K}$$
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