Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
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271.
A first order reaction is $$50\% $$ completed in $$1.26 \times {10^{14}}s.$$ How much time would it take for $$100\% $$ completion?
A
$$1.26 \times {10^{15}}s$$
B
$$2.52 \times {10^{14}}s$$
C
$$2.52 \times {10^{28}}s$$
D
$${\text{Infinite}}$$
Answer :
$${\text{Infinite}}$$
Reaction would be 100% complete only after infinite time which cannot be calculated.
272.
The rate of a reaction doubles when its temperature changes from $$300 K$$ to $$310 K.$$ Activation energy of such a reaction will be : $$\left( {R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}\,{\text{and}}\,\log 2 = 0.301} \right)$$
273.
For a reaction, $${A_2} + {B_2} \rightleftharpoons 2AB$$ the figure shows the path of the reaction in absence and presence of a catalyst. What will be the energy of activation for forward $$\left( {{E_f}} \right)$$ and backward $$\left( {{E_b}} \right)$$ reaction in presence of a catalyst and $$\Delta H$$ for the reaction? The dotted curve is the path of reaction in presence of a catalyst.
A
$${E_f} = 60\,kJ/mol,{E_b} = 70\,kJ/mol,$$ $$\Delta H = 20\,kJ/mol$$
B
$${E_f} = 20\,kJ/mol,{E_b} = 20\,kJ/mol,$$ $$\Delta H = 50\,kJ/mol$$
C
$${E_f} = 70\,kJ/mol,{E_b} = 20\,kJ/mol,$$ $$\Delta H = 10\,kJ/mol$$
D
$${E_f} = 10\,kJ/mol,{E_b} = 20\,kJ/mol,$$ $$\Delta H = - 10\,kJ/mol$$
274.
A radioactive isotope having a half - life period of 3 days was received after 12 days. If $$3g$$ of the isotope is left in the container, what would be the initial mass of the isotope?
275.
For the reaction, $$2{N_2}{O_5} \to 4N{O_2} + {O_2},$$ the rate
equation can be expressed in two ways $$ - \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = k\left[ {{N_2}{O_5}} \right]$$ and $$ + \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = k'\left[ {{N_2}{O_5}} \right]$$ $$k$$ and $$k'$$ are related as :
276.
The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduced to :
A
$$\frac{1}{2}g$$
B
$$\frac{1}{4}g$$
C
$$\frac{1}{8}g$$
D
$$\frac{1}{{16}}g$$
Answer :
$$\frac{1}{{16}}g$$
TIPS/Formulae :
$$N = {N_0}{\left( {\frac{1}{2}} \right)^n}$$
where, $$N=$$ Amount of radioactive substance which is left after certain number of half-life periods $$(n)$$
$${N_0} = $$ Initial amount of radioactive substance.
$$\eqalign{
& {\text{No}}{\text{. of half - lives = }}\frac{{{\text{total}}\,{\text{time}}}}{{{\text{half}}\,{\text{life}}\,{\text{period}}}} \cr
& = \frac{{560}}{{140}} \cr
& = 4 \cr} $$
In $$ānā$$ half-lives, the element will reduce to
$$\eqalign{
& {\left( {\frac{1}{2}} \right)^n} \times {\text{Initial}}\,{\text{wt}}{\text{.}} \cr
& {\text{ = }}{\left( {\frac{1}{2}} \right)^4} \times 1 \cr
& = \frac{1}{{16}}g \cr} $$
277.
The rate constant of a zero order reaction is $$2.0 \times {10^{ - 2}}mol\,{L^{ - 1}}{s^{ - 1}}.$$ If the concentration of the reactant after 25 seconds is $$0.5\,M.$$ What is the initial concentration?
A
0.5$$\,M$$
B
1.25$$\,M$$
C
12.5$$\,M$$
D
1.0$$\,M$$
Answer :
1.0$$\,M$$
$$\eqalign{
& {\text{For a zero order reaction}} \cr
& {\text{Rate constant}} = k = \frac{{a - x}}{t} \cr
& 2 \times {10^{ - 2}} = \frac{{a - 0.5}}{{25}} \cr
& a - 0.5 = 0.5 \cr
& a = 1.0M \cr} $$
278.
A first order reaction is half-completed in 45 minutes. How long does it need for $$99.9\% $$ of the reaction to be completed ?
280.
The rate equation for a reaction,
$${N_2}O \to {N_2} + \frac{1}{2}{O_2}$$
is Rate $$ = k{\left[ {{N_2}O} \right]^0} = k.$$ If the initial concentration of the reactant is $$a\,mol\,Li{t^{ - 1}},$$ the half-life period of the reaction is
A
$${t_{\frac{1}{2}}} = \frac{a}{{2k}}$$
B
$$ - {t_{\frac{1}{2}}} = ka$$
C
$${t_{\frac{1}{2}}} = \frac{a}{k}$$
D
$${t_{\frac{1}{2}}} = \frac{k}{a}$$
Answer :
$${t_{\frac{1}{2}}} = \frac{a}{{2k}}$$
For a zero order reaction
$${t_{\frac{1}{2}}} = \frac{a}{{2k}}$$