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After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is `1.155xx10^(-3)s^(-1)`?A. 600B. 100C. 60D. 10 |
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Answer» Correct Answer - A Rate constant `k=1.155xx10^(-3)s^(-1)` `k=(2.303)/(t)"log"(a)/((a-x))` `because" "a=a,(a-x)=(a)/(2)` `t_(1//2)=(2.303)/(k)"log"(a)/(a//2)=(2.303)/(1.155xx10^(-3))"log "2` `or" "t_(1//2)=600s` |
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