Equal volumes of two equimolar solutions of reactants A and B are mixed, andthe reaction A + B → C occurs. At the end of 1 hour, A is 90% reacted. How muchof A will be left unreacted at the end of 2 hours if the reaction is:a) First order in A and zero order in B.b) First order in A and first order in B.c) Zero order in both A and B.d) ) First order in A and one-half order in B.
A and B are joined by C. Ao is the initial concentration of A in a reaction mixture. Bo shows the initial concentration of B in the reaction mixture. The first order is A and the second order is B. k[A]/dt is the combination of d[A] and dt. ln [A] + -kt + ln Ao. after 1h, 90% A reacted, left 10% = 0.1 ln 0. 1Ao = -k(1) + ln Ao ln 0. 1 + ln Ao = -k + ln Ao => k = - ln 0.1 ln [A] = (ln 0. 1)t + ln Ao end of 2h, ln [A] = (ln 0. 1)(2) + ln Ao = ln 0. 01 + ln Ao = ln 0.01Ao 1% of A will remain unreacted at end of 2h. Zero order in both A and B. The rate of reaction is not dependent on the other variables. A and dt are related. Ao + -kt + Ao are related. end of 1h: 0.1Ao = -k + Ao => k= 0.9Ao [A] = (-0. 9Ao)t + Ao end of 2h: [A] = (-0. 9Ao)(2) + Ao = -0.8Ao reaction stop when A = 0, ie. All A has been reacting. Zero A present after 2h
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