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The rate constant of a reaction is given as `k=2.1xx10^(10)e^(-2700//RT)` It means that `"log "k " vs "1//T` will be a straight line with intercept on `"log "k" axis"=log2.1xx10^(10).` Number of effective collisions of temperature are `2.1xx10^(10)cm^(-3)s^(-1).` Half-life of a reaction increases of temperature. `"log"k" vs "1//T` will be a straight line with `slope=-(2700)/(2.303R).` Which of the above statements are true? Choose the correct option.A. I and IIB. II and IIIC. III and IVD. I and IV |
Answer» Correct Answer - d `k=2.1xx10^(10)e^(-2700//RT)` `" ""In " k="In "2.1xx10^(10)+"In "e^(-2700//RT)` `"log "k="log "2.1xx10^(10)-(2700)/(2.303RT)` `"log"k=-(2700)/(2.303RT)(1)/(T)+"log "2.1xx10^(10).` On comparing it with straight line equation, y=mx+c, we get `therefore" "m(slope)=-(2700)/(2.303R)` and `C("intercept")="log "2.1xx10^(10)` |
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