Order of Reaction

Q.

Mechanism of a hypothetical reaction $X_2+Y_2\to 2XY$ is given below
$I)X_2\rightleftharpoons X+X(fast\left)II\right)X+Y_2\to XY+Y\left(slow\right)III)X+Y\to XY(fast)$
The overall order of the reaction will be

1. 1

2. 0

3. 2

4. 1.5

Q. The conversion of molecules A to B follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by a factor of

1. $\frac{1}{4}$
2. $4$
3. $2$
4. $\frac{1}{2}$

Q. For the reaction 2A + 3B $\to$products, A is in excess and on changing the concentration of B from 0.1 M to 0.4 M, rate becomes doubled. Thus, rate law is

1. $\frac{dx}{dt}=k\left[A{]}^2\right[B{]}^2$
2. $\frac{dx}{dt}=k\left[A\right]\left[B\right]$
3. $\frac{dx}{dt}=k\left[A\right]0[B{]}^2$
4. $\frac{dx}{dt}=k[B{]}^{1/2}$

Q. Which one of the following is a second order reaction?

1. $H_2+B{r}_2\to 2HBr$
2. $C{H}_{3}COOC{H}_3+NaOH\to C{H}_{3}COONa+C{H}_{3}OH$
3. $N{H}_{4}N{O}_3\to N_2+3H_{2}O$
4. $H_2+C{l}_2\to 2HCl$

Q.

For the reaction : $aA+bB\to$ Products, the rate law is given by Rate = $k\left[A{]}^m\right[B{]}^n$. On making the concentration of A two-fold and halving that of B, the ratio of the new rate to the earlier rate of the reaction would be:

1. $n-1$

2. $m+n$

3. $2^{-(m+n)}$

4. $2^{(m-n)}$