1.

When in reaction concentration of reactant is 8 time increase then rate of reactant is doubled than what is order of reaction?

Answer»

First
`(1)/(3)`
0.5
0

Solution :Suppose the order =x
Initial rate =`r_(1)=k[R]^(x)`
New Rate =`r_(2)`=k`[8R]^(x)=K8^(x)[R]^(x)`
But new rate =`r_(2)=2r_(1)`
`therefore 2r_(1)=k 8^(x)[R]^(x)`
So, `(2r_(1))/(r_(1))=(k(8)^(x)[R]^(x))/(k[R]^(x))`
`therefore 2=8^(x)`
`therefore 2=8^((1)/(3))` So,order of reaction`x=(1)/(3)`


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