JEE MAIN - Chemistry (2021 - 17th March Morning Shift - No. 18)
For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ $$\times$$ 10$$-$$3 s$$-$$1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]
Answer
2
Explanation
$$k = {1 \over t}\ln \left[ {{a \over {a - x}}} \right]$$
$$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$$
$$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$$
$$k = {{2.303} \over {570}} \times 0.5$$
$$k = 2 \times {10^{ - 3}}{s^{ - 1}}$$
$$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$$
$$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$$
$$k = {{2.303} \over {570}} \times 0.5$$
$$k = 2 \times {10^{ - 3}}{s^{ - 1}}$$
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