Question

A and B are two different chemical species undergoing 1st order decomposition with half lives equal to $$5$$ sec. and $$7.5$$ sec. respectively. If the initial concentration of A and B are in the ratio $$3:2$$, calculate $$\cfrac{{C}_{A}}{{C}_{B}}$$, after three half lives of A. Report your answer after multiplying it with $$100$$.

Solution

$${ t }_{ 1/2 }(A)=5sec=\cfrac { 0.693 }{ { k }_{ A } } \\ { k }_{ A }=0.1386\quad { sec }^{ -1 }\\ { t }_{ 1/2 }(B)=1.5sec=\cfrac { 0.693 }{ { k }_{ B } } \\ { k }_{ B }=0.0924\quad { sec }^{ -1 }$$ Let initial concentration of A be  150. Then initial concentration of B is 100. Three half lives of $$A=5\times 3=15sec.\\ { C }_{ A }=\cfrac { 150 }{ { 2 }^{ 3 } } =18.75\quad mol/L$$$$15sec=2$$ half lives of B$${ C }_{ B }=\cfrac { 100 }{ { 2 }^{ 2 } } =25\quad mol/L\\ \cfrac { { C }_{ A } }{ { C }_{ B } } =\cfrac { 18.75 }{ 25 } =0.75$$ Multiplying it by 100 we get$$0.75\times 100=75$$Chemistry

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