1.

What will be the initial rate of reaction if its rate constant is 10^(-3)s^(-1) and the concentration of the reactant is 0.2 mol L^(-1)? What fraction of the reactant will be converted into the products in 200 seconds ?

Answer»

SOLUTION :INITIAL rate `=K xx [A]`
Substituting the values, we get
Initial rate `=10^(-3)s^(-1) xx 0.2 "mol L"^(-1)=2xx10^(-4)"mol L"^(-1)s^(-1)`.
To obtain fraction of reactants that will be CONVERTED into products after 200 s, proceed as follows:
`t=(2.303)/(k)"log"(1)/(N)`
or `"log"(1)/(N)=(200sxx10^(-3)s^(-1))/(2.303) =(2xx10^(-1))/(2.303)=0.086`
or `(1)/(N)="Antilog of"0.086=1.219`
or `N=(1)/(1.219)=0.820 or 82%`
Per cent that will change into products `=100-82=18%`.


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