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At 27^(@)C in the presence of a catalyst, the activation energy of a reaction is lowered by 2 kcal. Calculate by how much the rate of reaction will increase ? |
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Answer» Solution :In the ABSENCE of catalyst, SUPPOSE rate constant = k. Then `logk=logA-(E_(a))/(2.303"RT")""…(i)` In the presence of catalyst, suppose rate constant = k'. Now, activation energy `=E_(a)-2("if "E_(a)" is in KCAL "mol^(-1))` `:.logk'=logA-(E_(a)-2)/(2.303"RT")" or "logk'=logA-(E_(a))/(2.303"RT")+(2"kcal mol"^(-1))/(2.303"RT")""...(ii)` Subtracting eqn (i) from eqn (ii), we get `logk'-logk=(2" kcal mol"^(-1))/(2.303"RT")` `log""(k')/(k)=(2" kcal mol"^(-1))/(2.303(2xx10^(-3)"kcal K"^(-1)mol^(-1))(300" K"))=1.4474""(R=2xx10^(-3)"kcal K"^(-1)mol^(-1))` or `(k')/(k)="Antilog "1.4474=28" or "k'=28" k`, i.e., the rate of REACTION will INCREASE 28 times. |
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