Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

For a zero order, rate of reaction does not change with time.

$$_0{n^1}{ \to _{ + 1}}{P^1}{ + _{ - 1}}{e^0}$$

As the concentration of reactants increases the probability of collisions increases hence collision frequency increases.

$$r = k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}.$$    When the volume is reduced to $$\frac{1}{2}$$ , the conc. will double
$$\therefore \,\,{\text{New}}\,{\text{rate}} = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8k\left[ {{O^2}} \right]{\left[ {NO} \right]^2}$$
The new rate increases to eight times of its initial.

More activations energy, slow reaction rate.

Rate constant of a reaction does not depend upon concentrations of the reactants. Thus, it will remain same.

\[_{90}^{234}Th\xrightarrow{{ - \beta }}\,_{91}^{234}X\xrightarrow{{ - \beta }}\,_{92}^{234}Th\xrightarrow{{ - \alpha }}\,_{90}^{230}Th\]

$$\eqalign{ & {}_{92}{U^{238}}{ \to _{90}}{U^{234}} + _2^4He \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\alpha - {\text{particle}}} \right) \cr} $$

$$\eqalign{ & {t_{\frac{1}{2}}} = 3\,hrs.\,T = 18\,hours\,\,\,\because T = n \times {t_{\frac{1}{2}}} \cr & \therefore n = \frac{{18}}{3} = 6 \cr & {\text{Initial mass}}\left( {{C_0}} \right) = 256\,g \cr & \therefore \,{C_n} = \frac{{{C_0}}}{{{2^n}}} = \frac{{256}}{{{{\left( 2 \right)}^6}}} = \frac{{256}}{{64}} = 4g. \cr} $$

According to rate law expression, $${\text{Rate}} \propto \left[ R \right]$$
Thus, rate of a reaction decreases with passage of time as the concentration of reactants decreases.