Assuming that 2NO + O_(2) to 2NO_(2) is a single-step reaction, what will be the rate of reaction when the volume of the reaction vessel is reduced of 1//4^(th) of the initial value ? The original rate of reaction is 64 mol/L/s.

Assuming that 2NO + O_(2) to 2NO_(2) is a single-step reaction, what w

1.	Assuming that 2NO + O_(2) to 2NO_(2) is a single-step reaction, what will be the rate of reaction when the volume of the reaction vessel is reduced of 1//4^(th) of the initial value ? The original rate of reaction is 64 mol/L/s.
Answer» Solution :`2NO + O_(2)to 2NO_(2)` The rate of the reactioin r`=k[NO]^(2)[O_(2)]`, as it is a single-step reaction. When the volume of the reaction vessel is reduced to `1//4^(TH)` of the initial volume, the concentration of each reactant is increased by four-times. Since the reaction is a second order with respect to NO and first order with respect to `O_(2)` , the rate of reaction INCREASES by 16-times for NO and 4-times for `O_(2)`. On the WHOLE , the rate of reaction increases by 64-times . The ORIGINAL rate of reaction is 64 mol/L/s and the new rate of reaction is 64-times more than this. `r_(2)=64xx64=4096=4.096xx10^(3)` mol/L/s.