$${t_{\frac{1}{2}}} = 5\,{\text{years,}}\,T = 15\,{\text{years}}$$ hence total number of half life periods $${\text{ = }}\frac{{15}}{5} = 3$$
∴ Amount left $$ = \frac{{64}}{{{{\left( 2 \right)}^3}}} = 8g$$
122.
A first order reaction has a rate constant $$1.15 \times {10^{ - 3}}\,{s^{ - 1}}.$$ How long will $$5\,g$$ of this reactant take to reduce to $$3\,g?$$
123.
In a reaction, $$2X \to Y,$$ the concentration of $$X$$ decreases from $$3.0\,moles{\text{/litre}}$$ to $${\text{2}}{\text{.0}}\,moles{\text{/litre}}$$ in 5 minutes. The rate of reaction is
124.
reaction which is of first order $$w.r.t.$$ reactant $$A,$$ has a rate constant \[6\,{{\min }^{-1}}.\] If we start with $$\left[ A \right] = 0.5\,mol\,{L^{ - 1}},$$ when would $$\left[ A \right]$$ reach the value of $$0.5\,mol\,{L^{ - 1}}$$
125.
The rate constant of a first order reaction is $$15 \times {10^{ - 3}}\,{s^{ - 1}}.$$ How long will $$5.0\,g$$ of this reactant take to reduce to $$3.0\,g?$$
126.
Compounds $$'A'$$ and $$'B'$$ react according to the following chemical equation.
$${A_{\left( g \right)}} + 2{B_{\left( g \right)}} \to 2{C_{\left( g \right)}}$$
Concentration of either $$'A'$$ or $$'B'$$ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
Experiment
Initial concentration of $$\left[ A \right]/mol\,{L^{ - 1}}$$
Initial concentration of $$\left[ B \right]/mol\,{L^{ - 1}}$$
Initial rate of formation of $$\left[ C \right]/mol\,{L^{ - 1}}\,{s^{ - 1}}$$
1.
0.30
0.30
0.10
2.
0.30
0.60
0.40
3.
0.60
0.30
0.20
A
$${\text{Rate}} = k{\left[ A \right]^2}\left[ B \right]$$
B
$${\text{Rate}} = k\left[ A \right]{\left[ B \right]^2}$$
C
$${\text{Rate}} = k\left[ A \right]\left[ B \right]$$
D
$${\text{Rate}} = k{\left[ A \right]^2}{\left[ B \right]^0}$$
Answer :
$${\text{Rate}} = k\left[ A \right]{\left[ B \right]^2}$$
Rate of reaction $$ \propto $$ cone. of reactants
As the reaction proceeds, concentration of the reactants decreases hence the rate also keeps on decreasing with time.
128.
For the reaction $${N_2} + 3{H_2} \to 2N{H_3},$$ how are the rate of reaction expressions inter-related $$\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$ and $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}}?$$
$${H_2}$$ decreases three times as fast as that of $${N_2}$$ while $$N{H_3}$$ increases twice as fast as that of $${N_2}$$ decreases.
Hence, Rate $$ = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
129.
The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at $${25^ \circ }C$$ are $$3.0 \times {10^{ - 4}}{s^{ - 1}},$$ $$104.4\,kJ\,mo{l^{ - 1}}$$ and $$6.0 \times {10^{14}}{s^{ - 1}}$$ respectively. The value of the rate constant as $$T \to \infty $$ is,
The Arrhenius equation is $${\text{:}}\,\,k = A\,\exp \left( { - {E_a}/RT} \right)$$
as $$T \to \infty ,\,\exp \left( { - {E_a}/RT} \right) \to 1.$$ Hence, $$k=A$$
where $$A,$$ the Arrhenius parameter is $$6.0 \times {10^{14}}{s^{ - 1}}$$
[ NOTE : $$'A'$$ is also known as frequency factor ]
130.
Which of the following nuclear reactions will generate an isotope?
NOTE : Isotopes are atoms of same element having same atomic number but different atomic masses. Neutron has atomic number 0 and atomic mass 1. So loss of neutron will generate isotope. e.g.,
$$_{92}{U^{238}}{ + _0}{n^1}{ \to _{92}}{U^{239}}$$