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Rate constant for a reaction is `10^(-3) S^(-1)`. How much time is required to reduce the initial concentration of reactant to 25%A. 693 sB. 1386 sC. 6930 sD. 2029 s |
Answer» Correct Answer - B `t=(2.303)/(K)"log"(a)/(a-x)` where," "k=rate constant=`10^(-3)s^(-1)` a=I"initial amount "=100 a-x= amount left after time, t =25 t=time required to reduce the initial concentration of reactant to 25% reaction `therefore " " t=(2.303)/(10^(-3))"log"(100)/(25)=(2.303)/(10^(-3))`"log" 4 =` (2.303xx0.6020)/(10^(-3))=1386s` |
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