Question

The rate law for the reaction between the substance a and b is given by rate is equal to k into a power n b power m and doubling the concentration of a and half the concentration of b the ratio of new rate of the earlier rate of the reaction will be

Answer 1

Answer:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A+B\rightarrow products

rate =k[A]^n[B]^m

rate =k\frac{C_a}{V}^n\times \frac{C_b}{V}^m

On doubling the concentration of A, making the volume of B half.

rate'=k\frac{2C_a}{V}^n\times \frac{C_b}{2V}^m

rate'=k\frac{2^nC_a^n}{V^n}\times \frac{C_b^m}{2^mV^m}

Thus rate'=2^{n-m}rate

\frac{rate'}{rate}=2^{n-m}

Answer 2

$\huge\rm\underline{\underline{\pink{Answer}}}$

It measures the work done n

per unit charger.

It is defined as the difference in

electric potential between two points in an electric field,equal to the work

done per unit quantity of charge in moving it from one point to another

in an electrostatic field