A certain substance X decomposes. Fifty percent of X remains
aft...

Question

A certain substance X decomposes. Fifty percent of X remains 
after 100 minutes. How much X remains after 200 minutes if 
the reaction order with respect to X is (c) second order?

Accepted Answer

Hi everyone here we have a question that tells us in an experiment, 1.2 molar of compound B undergoes decomposition after 120 minutes. Only 50% of the remains determine the time it takes for 90% of B to decompose. If the decomposition is in first order, so at T Equals 120 minutes, 50% remains. So at T 1/ Equals 120 minutes for forced order reaction. T one half Equals 0.693, divided by K. Solving for K. K equals 0. Over 120 minutes Equals 5.77, 5 times 10 to the negative third minutes, inverse. The integrated rate law. For first order, the natural log of the concentration of a T equals negative Katie plus the natural log of the concentration of a zero. So the natural log of 0. Times 1.2 Moller equals negative 5.775 times 10 to the negative third inverse minutes. T plus the natural log of 1.2 Molar, So negative 2.120 equals negative 5.77, 5 Times 10 to the negative 3rd inverse minute T plus 0.182. So t equals 398.61 minutes. And that is our final answer. Thank you for watching. Bye.

A certain substance X decomposes. Fifty percent of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is (c) second order?

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