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Chemical Kinetics

NEET Chemistry · 91 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
Activation energy (E aa ) and rate constants (k 1 and k 2 ) of a chemical reaction at two different temperatures (T 1 and T 2 ) are related by
A ln⁡k2k1=−EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)
B ln⁡k2k1=−EaR(1T2−1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)
C ln⁡k2k1=−EaR(1T2+1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} + {1 \over {{T_1}}}} \right)
D ln⁡k2k1=EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)
Correct Answer
Option B
Solution

Let

k1=Ae−EaRT1{k_1} = A{e^{ - {{{E_a}} \over {R{T_1}}}}}
ln⁡k1=ln⁡A−EaRT1\ln {k_1} = \ln A - {{{E_a}} \over {R{T_1}}}

......(1)

k2=Ae−EaRT2{k_2} = A{e^{ - {{{E_a}} \over {R{T_2}}}}}
ln⁡k2=ln⁡A−EaRT2\ln {k_2} = \ln A - {{{E_a}} \over {R{T_2}}}

....(2) From eq.(1) and (2), we have

ln⁡k2−ln⁡k1=ln⁡A−EaRT2−ln⁡A+EaRT1\ln {k_2} - \ln {k_1} = \ln A - {{{E_a}} \over {R{T_2}}} - \ln A + {{{E_a}} \over {R{T_1}}}

⇒\Rightarrow

ln⁡k2k1=EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)

⇒\Rightarrow

ln⁡k2k1=−EaR(1T2−1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)
Q62
β\beta - particle is emitted in radioactivity by
A conversion of proton to neutron
B from outermost orbit
C conversion of neutron to proton
D β\beta -particle is not emitted
Correct Answer
Option C
Solution
0n1→+1p1+−1e{}_0{n^1} \to {}_{ + 1}{p^1} + {}_{ - 1}{e^{}}
Q63
The rate of a chemical reaction doubles for every 10oC rise of temperature. If the temperature is raised by 50oC , the rate of the reaction increases by about :
A 24 times
B 32 times
C 64 times
D 10 times
Correct Answer
Option B
Solution

Since for every

10∘C{10^ \circ }C

rise in temperature rate doubles for

50∘C{50^ \circ }C

rise in temp increase in reaction rate

=25=32= {2^5} = 32

times

Q64
In the transformation of 92238U{}_{92}^{238}U to 92234U{}_{92}^{234}U, if one emission is an α-particle, what should be the other emission(s)?
A Two β−\beta^-
B Two β−\beta^- and one β+\beta^+
C One β−\beta^- and one γ\gamma
D One β+\beta^+ and One β−\beta^-
Correct Answer
Option A
Solution
92238U⟶−α90234Th⟶ −2β92234U{}_{92}^{238}U\overset{{ - \alpha }}\longrightarrow {}_{90}^{234}Th\overset{{\, - 2\beta }}\longrightarrow {}_{92}^{234}U
Q65
The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of it remaining undecayed after 18 hours would be
A 8.0 g
B 12.0 g
C 16.0 g
D 4.0 g
Correct Answer
Option D
Solution
t1/2=3 {t_{1/2}} = 3\,

hrs.

T=18T=18

hours as

    T=n×t1/2\,\,\,\,T = n \times {t_{1/2}}

∴\therefore

    n=183=6\,\,\,\,n = {{18} \over 3} = 6

Initial mass

(C0)=256 g\left( {{C_0}} \right) = 256\,g

∴\therefore

    Cn=C02n=256(2)6=25664=4g.\,\,\,\,{C_n} = {{{C_0}} \over {{2^n}}} = {{256} \over {{{\left( 2 \right)}^6}}} = {{256} \over {64}} = 4g.
Q66
The rate law for the reaction below is given by the expression k [A] [B] A + B → \to Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :
A k
B k/3
C 3k
D 9k
Correct Answer
Option A
Solution

Rate constant only depends on temperature only and it is independed of concentration of reactants.

Q67
It is true that :
A A first order reaction is always a single step reaction
B A zero order reaction is a multistep reaction
C A zero order reaction is a single step reaction
D A second order reaction is always a multistep reaction
Correct Answer
Option B
Solution

Zero order reaction has complex mechanism. Zero order reaction is a multistep reaction.

Q68
The differential rate law for the reaction H2 + I2 →\to 2HI is
A −d[H2]dt - {{d\left[ {{H_2}} \right]} \over {dt}} = −d[I2]dt - {{d\left[ {{I_2}} \right]} \over {dt}} = −d[HI]dt - {{d\left[ {{HI}} \right]} \over {dt}}
B d[H2]dt {{d\left[ {{H_2}} \right]} \over {dt}} = d[I2]dt {{d\left[ {{I_2}} \right]} \over {dt}} = d[HI]dt {{d\left[ {{HI}} \right]} \over {dt}}
C 12d[H2]dt{1 \over 2}{{d\left[ {{H_2}} \right]} \over {dt}} = 12d[I2]dt{1 \over 2}{{d\left[ {{I_2}} \right]} \over {dt}} = −d[HI]dt - {{d\left[ {{HI}} \right]} \over {dt}}
D −2d[H2]dt - 2{{d\left[ {{H_2}} \right]} \over {dt}} = −2d[I2]dt - 2{{d\left[ {{I_2}} \right]} \over {dt}} = d[HI]dt{{d\left[ {{HI}} \right]} \over {dt}}
Correct Answer
Option D
Solution

rate of appearance of

HI=12d[HI]dtHI = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}

rate of formation of

H2=−d[H2]dt{H_2} = {{ - d\left[ {{H_2}} \right]} \over {dt}}

rate of formation of

I2=−d[I2]dt{I_2} = {{ - d\left[ {{I_2}} \right]} \over {dt}}

hence

−d[H2]dt=−−d[I2]dt=12d[HI]dt{{ - d\left[ {{H_2}} \right]} \over {dt}} = - {{ - d\left[ {{I_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}

or

    \,\,\,\,
−2d[H2]dt=−2d[I2]dt=d[HI]dt- {{2d\left[ {{H_2}} \right]} \over {dt}} = - {{2d\left[ {{I_2}} \right]} \over {dt}} = {{d\left[ {HI} \right]} \over {dt}}
Q69
For a reaction of order n, the unit of the rate constant is :
A mol1−-n L1−-n s
B mol1−-n L2n s−-1
C mol1−-n Ln−-1 s−-1
D mol1−-n L1−-n s−-1
Correct Answer
Option C
Solution

Rate = k[A]n comparing units

(mol/l)sec⁡=k(moll)n{{(mol/l)} \over {\sec }} = k{\left( {{{mol} \over l}} \right)^n}
⇒k=mol(l−n)l(n−1)s−1\Rightarrow k = mo{l^{(l - n)}}{l^{(n - 1)}}{s^{ - 1}}
Q70
For a chemical reaction A+B→\mathrm{A}+\mathrm{B} \rightarrow Product, the order is 1 with respect to A\mathrm{A} and B\mathrm{B}. .tg .tg Rate\mathrm{Rate} mol L−1 S−1\mathrm{mol~L^{-1}~S^{-1}} [A]\mathrm{[A]} mol L−1\mathrm{mol~L^{-1}} [B]\mathrm{[B]} mol L−1\mathrm{mol~L^{-1}} 0.10 20 0.5 0.40 xx 0.5 0.80 40 yy What is the value of xx and yy ?
A 80 and 4
B 160 and 4
C 80 and 2
D 40 and 4
Correct Answer
Option C
Solution
r=K[A]1[ B]10.1=K(20)1(0.5)1........(i)0.40=K(x)1(0.5)1........(ii)0.80=K(40)1(y)1........(iii)\begin{aligned} & \mathrm{r}=\mathrm{K}[\mathrm{A}]^1[\mathrm{~B}]^1 \\\\ & 0.1=\mathrm{K}(20)^1(0.5)^1 ........(i) \\\\ & 0.40=\mathrm{K}(\mathrm{x})^1(0.5)^1 ........(ii) \\\\ & 0.80=\mathrm{K}(40)^1(\mathrm{y})^1 ........(iii) \end{aligned}

From (i) and (ii)

x=80x=80

From (i) and (iii)

y=2y=2
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