For the reaction A + B `to` products, rate low expression for rate =`k[A][B]^(2)`. If the volume of the vessel is reduced to `1/3` of its original volume then what will be the effect on the rate of reaction?
rate = `k[A][B]^(2)`
By reducing the volume to `1/3` of its original volume, the concentration of the reactants will become three times.
`therefore` rate = `k[3A][3B]^(2)=27k[A][B]^(2)` ltbr. Hence, the reaction rate will become 27 times its initial rate.
a) The rate law epression is:
Rate(r) = `k[A]^(1//3)[B]^(2)therefore` Order of reaction `=1/3 + 2=2(1/3)`
b) Rate constant, k =`("rate")/([A]^(1//3)[B]^(2))`
`therefore` Units of rate constant = `(moldm^(-3)s^(-1))/((mol dm^(-3^(1//3))(mol dm^(-3))^(2)))`
`=(moldm^(-3))^(1-7//3)s^(-1)=(mol...
In an elementary reaction, the number of atoms or ions which are colliding to react is known as molecularity. Had the reaction been of elementary nature (single step reaction) the...
Step i. Calculation of the rate constant.
Rate constant(k)`=(0.693)/(t_(1//20))= (0.693)/(3.33 "hours")=0.2081 hr^(-1)`
Step II. Calculation of amount of A left unreacted
`k=0.2081 hr^(-1)`, a=1mol, t=9 hours.
`k=2.303/t log a/(a-x)`
`log...