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Chemical Kinetics

NEET Chemistry · 91 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
For a certain reaction R→R \rightarrow Product, the plot of concentration [R][R] vs time has a negative slope as shown. The order of reaction is :
A 0
B 1
C 2
D 2.5
Correct Answer
Option A
Solution

For zero order

[R]=[R0]−kt[R]=\left[R_0\right]-k t

So -ve slope shows zero order reaction

Q2
If the rate constant of a reaction is 0.03 s−10.03 \mathrm{~s}^{-1}, how much time does it take for 7.2 mol L−17.2 \mathrm{~mol} \mathrm{~L}^{-1} concentration of the reactant to get reduced to 0.9 mol L−10.9 \mathrm{~mol} \mathrm{~L}^{-1} ? (Given: log⁡2=0.301\log 2=0.301 )
A 210 s
B 21.0 s
C 69.3 s
D 23.1 s
Correct Answer
Option C
Solution

To determine how long it takes for the concentration of a reactant to decrease from 7.2 mol L−1^{-1} to 0.9 mol L−1^{-1}, we use the formula for the first-order reaction: t=2.303klog⁡aa−x t = \dfrac{2.303}{k} \log \dfrac{a}{a-x} Where: k=0.03 s−1 k = 0.03 \, \text{s}^{-1} is the rate constant. a=7.2 mol L−1 a = 7.2 \, \text{mol L}^{-1} is the initial concentration. a−x=0.9 mol L−1 a-x = 0.9 \, \text{mol L}^{-1} is the final concentration.

Plug these values into the equation: t=2.3030.03log⁡7.20.9=2.3030.03log⁡8=2.3030.03×3×log⁡2=2.3030.03×3×0.301=69.3 s \begin{aligned} t & = \dfrac{2.303}{0.03} \log \dfrac{7.2}{0.9} \\ & = \dfrac{2.303}{0.03} \log 8 \\ & = \dfrac{2.303}{0.03} \times 3 \times \log 2 \\ & = \dfrac{2.303}{0.03} \times 3 \times 0.301 \\ & = 69.3 \, \text{s} \end{aligned} Therefore, it takes 69.3 seconds for the concentration to decrease from 7.2 mol L−1^{-1} to 0.9 mol L−1^{-1}.

Q3
If the half-life ( t1/2t_{1 / 2} ) for a first order reaction is 1 minute, then the time required for 99.9%99.9 \% completion of the reaction is closest to :
A 5 minutes
B 10 minutes
C 2 minutes
D 4 minutes
Correct Answer
Option B
Solution

For a first‐order reaction the half‐life and rate constant are related by

k=ln⁡2t1/2=0.6931  min=0.693  min−1.k=\frac{\ln2}{t_{1/2}}=\frac{0.693}{1\;\text{min}}=0.693\;\text{min}^{-1}.

We want 99.9% completion, so the fraction remaining is 0.001. Using

[A]t[A]0=e−kt=0.001\frac{[A]_t}{[A]_0}=e^{-kt}=0.001

take natural logs:

−kt=ln⁡(0.001)=−6.9078-kt=\ln(0.001)=-6.9078

hence

t=6.90780.693≈9.97  min≈10  min.t=\frac{6.9078}{0.693}\approx9.97\;\text{min}\approx10\;\text{min}.

Answer: Option B, 10 minutes.

Q4
<p>Following data is for a reaction between reactants A and B :</p> <p> <table class=tg style=undefined;table-layout: fixed; width: 580px><colgroup> <col style=width: 245px> <col style=width: 94px> <col style=width: 241px> </colgroup> <thead> <tr> <th class=tg-7btt>Rate <br>mol L−1 s−1\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}</th> <th class=tg-7btt>[A]\mathrm{[A]}</th> <th class=tg-7btt>[B]\mathrm{[B]}</th> </tr></thead> <tbody> <tr> <td class=tg-c3ow><br>2×10−3<br><br>2 \times 10^{-3}<br></td> <td class=tg-c3ow>0.1 M</td> <td class=tg-c3ow>0.1 M</td> </tr> <tr> <td class=tg-c3ow><br>4×10−3<br><br>4 \times 10^{-3}<br></td> <td class=tg-c3ow>0.2 M</td> <td class=tg-c3ow>0.1 M</td> </tr> <tr> <td class=tg-c3ow><br>1.6×10−2<br><br>1.6 \times 10^{-2}<br></td> <td class=tg-c3ow>0.2 M</td> <td class=tg-c3ow>0.2 M</td> </tr> </tbody> </table></p> <p> The order of the reaction with respect to A and B, respectively, are  \text{ The order of the reaction with respect to } \mathrm{A} \text{ and } \mathrm{B} \text{, respectively, are } </p>
A 1, 0
B 0, 1
C 1, 2
D 2, 1
Correct Answer
Option C
Solution

Let the rate equation is

 Rate =k[A]×[B]y\text{ Rate }=k[A] \times[B]^y

Therefore, we can write

2×10−3=k[0.1]x[0.1]y..... (i)4×10−3=k[0.2]x[0.1]y..... (ii)1.6×10−2=k[0.2]x[0.2]y..... (iii)\begin{aligned} & 2 \times 10^{-3}=k[0.1]^x[0.1]^y \quad \text{..... (i)}\\ & 4 \times 10^{-3}=k[0.2]^x[0.1]^y \quad \text{..... (ii)}\\ & 1.6 \times 10^{-2}=k[0.2]^x[0.2]^y \quad \text{..... (iii)} \end{aligned}
 (ii) ÷ (i); 4×10−32×10−3=k[0.2]x[0.1]yk[0.1]x[0.1]y⇒21=(0.2)x(0.1)x=(21)x∴x=1\begin{aligned} & \text{ (ii) } \div \text{ (i); } \\ & \frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y} \\ & \Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x \\ & \therefore \quad x=1 \end{aligned}
 (ii) ÷ (iii); 4×10−31.6×10−2=k[0.2]x[0.1]yk[0.2]x[0.2]y⇒14=(0.1)y(0.2)y=(12)y∴y=2∴ Rate =k[A]1[B]2\begin{aligned} & \text{ (ii) } \div \text{ (iii); } \\ & \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^x[0.1]^y}{k[0.2]^x[0.2]^y} \\ & \Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y \\ & \therefore \quad y=2 \\ & \therefore \quad \text{ Rate }=k[A]^1[B]^2 \end{aligned}

First order with respect to A while second order with respect to B.

Q5
Rate constants of a reaction at 500 K500 \mathrm{~K} and 700 K700 \mathrm{~K} are 0.04 s−10.04 \mathrm{~s}^{-1} and 0.14 s−10.14 \mathrm{~s}^{-1}, respectively; then, activation energy of the reaction is : (Given: log⁡3.5=0.5441,R=8.31 J K−1 mol−1\log 3.5=0.5441, \mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1})
A 182310 J
B 18500 J
C 18219 J
D 18030 J
Correct Answer
Option C
Solution
K=Ae−Ea/RTK=A e^{-E_a / R T}

After taking In both side

ln⁡K=ln⁡A−EaRT\ln \mathrm{K}=\ln \mathrm{A}-\frac{E_a}{R T}
ln⁡K1=ln⁡A−EaRT1\ln K_1=\ln A-\frac{E_a}{R T_1}

at temp.

T1T_1

.... (i)

ln⁡K2=ln⁡A−EaRT2\ln K_2=\ln A-\frac{E_a}{R T_2}

at temp.

T2T_2

.... (ii)

 (ii) − (i) ln⁡K2−ln⁡K1=EaR[1T1−1T2]ln⁡K2K1=EaR[1500−1700]ln⁡0.140.04=EaR[700−500500×700]ln⁡144=EaR[200500×700]log⁡3.5=Ea2.303×R[1250×7]0.5441=Ea2.303×8.31[1250×7]Ea=0.5441×8.31×250×7×2.303=0.5441×83.1×25×7×2.303=18222.65≈18219 J\begin{aligned} & \text{ (ii) }- \text{ (i) } \\ & \ln K_2-\ln K_1=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ & \ln \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{1}{500}-\frac{1}{700}\right] \\ & \ln \frac{0.14}{0.04}=\frac{E_a}{R}\left[\frac{700-500}{500 \times 700}\right] \\ & \ln \frac{14}{4}=\frac{E_a}{R}\left[\frac{200}{500 \times 700}\right] \\ & \log 3.5=\frac{E_a}{2.303 \times R}\left[\frac{1}{250 \times 7}\right] \\ & 0.5441=\frac{E_a}{2.303 \times 8.31}\left[\frac{1}{250 \times 7}\right] \\ & E_a=0.5441 \times 8.31 \times 250 \times 7 \times 2.303 \\ & =0.5441 \times 83.1 \times 25 \times 7 \times 2.303 \\ & =18222.65 \\ & \approx 18219 \mathrm{~J} \end{aligned}
Q6
Activation energy of any chemical reaction can be calculated if one knows the value of
A rate constant at standard temperature
B probability of collision
C orientation of reactant molecules during collision
D rate constant at two different temperatures
Correct Answer
Option D
Solution

To calculate value of

Ea\mathrm{E_a}

Equation used is

log⁡(k2k1)=Ea2.303R(1T1−1T2)\mathrm{\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}

Hence

Ea\mathrm{E_a}

can be calculated if value of rate constant k is known at two different temperatures

T1\mathrm{T_1}

and

T2\mathrm{T}_2

.

Q7
The rate of a reaction quadruples when temperature changes from 27∘C27^{\circ} \mathrm{C} to 57∘C57^{\circ} \mathrm{C}. Calculate the energy of activation. Given R=8.314 J K−1 mol−1,log⁡4=0.6021\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021
A 38.04 kJ/mol38.04 \mathrm{~kJ} / \mathrm{mol}
B 380.4 kJ/mol380.4 \mathrm{~kJ} / \mathrm{mol}
C 3.80 kJ/mol3.80 \mathrm{~kJ} / \mathrm{mol}
D 3804 kJ/mol3804 \mathrm{~kJ} / \mathrm{mol}
Correct Answer
Option A
Solution

To solve this problem, we will use the Arrhenius equation, which relates the rate constant (

kk

) of a chemical reaction to the temperature (

TT

) and the activation energy (

EaE_a

). The equation in its logarithmic form is:

log⁡k=log⁡A−Ea2.303RT\log k = \log A - \frac{E_a}{2.303RT}

where:

kk

is the rate constant,

AA

is the pre-exponential factor (frequency factor),

EaE_a

is the activation energy,

RR

is the universal gas constant (

8.314 J mol−1K−18.314 \, \text{J mol}^{-1} \text{K}^{-1}

),

TT

is the temperature in Kelvin,

2.3032.303

is the factor to convert from natural log to common log. According to the problem, the rate of the reaction quadruples (

k2=4k1k_2 = 4k_1

) when the temperature increases from

27∘C27^{\circ}C

(which equals

27+273.15=300.15 K27 + 273.15 = 300.15 \, K

) to

57∘C57^{\circ}C

(which equals

57+273.15=330.15 K57 + 273.15 = 330.15 \, K

). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have:

log⁡k1=log⁡A−Ea2.303R×300.15\log k_1 = \log A - \frac{E_a}{2.303R \times 300.15}
log⁡k2=log⁡A−Ea2.303R×330.15\log k_2 = \log A - \frac{E_a}{2.303R \times 330.15}

With

k2=4k1k_2 = 4k_1

, substituting in the values and taking their difference:

log⁡4k1−log⁡k1=(log⁡A−Ea2.303×8.314×330.15)−(log⁡A−Ea2.303×8.314×300.15)\log 4k_1 - \log k_1 = \left(\log A - \frac{E_a}{2.303 \times 8.314 \times 330.15}\right) - \left(\log A - \frac{E_a}{2.303 \times 8.314 \times 300.15}\right)
log⁡4=Ea2.303×8.314(1300.15−1330.15)\log 4 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300.15} - \frac{1}{330.15}\right)

Given that

log⁡4=0.6021\log 4 = 0.6021

, we now solve for

EaE_a

:

0.6021=Ea2.303×8.314(1300.15−1330.15)0.6021 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300.15} - \frac{1}{330.15}\right)

First, calculate

1300.15−1330.15\frac{1}{300.15} - \frac{1}{330.15}

:

1300.15≈0.003332and1330.15≈0.003029\frac{1}{300.15} \approx 0.003332 \quad \text{and} \quad \frac{1}{330.15} \approx 0.003029
1300.15−1330.15=0.003332−0.003029=0.000303\frac{1}{300.15} - \frac{1}{330.15} = 0.003332 - 0.003029 = 0.000303

So,

0.6021=Ea19.141872×0.0003030.6021 = \frac{E_a}{19.141872} \times 0.000303
Ea19.141872=0.60210.000303=1987.05\frac{E_a}{19.141872} = \frac{0.6021}{0.000303} = 1987.05

Solve for

EaE_a

:

Ea=1987.05×19.141872=38067 J/mol≈38.067 kJ/molE_a = 1987.05 \times 19.141872 = 38067 \, \text{J/mol} \approx 38.067 \, \text{kJ/mol}

Therefore, the activation energy

EaE_a

is

38.07 kJ/mol38.07 \, \text{kJ/mol}

, which best matches Option A:

38.04 kJ/mol38.04 \, \text{kJ/mol}
Q8
For a reaction 3 A→2 B3 \mathrm{~A} \rightarrow 2 \mathrm{~B} The average rate of appearance of B\mathrm{B} is given by Δ[B]Δt\dfrac{\Delta[B]}{\Delta t}. The correct relation between the average rate of appearance of B\mathrm{B} with the average rate of disappearance of A is given in option :
A −Δ[A]Δt\dfrac{-\Delta[\mathrm{A}]}{\Delta \mathrm{t}}
B −3Δ[A]2Δt\dfrac{-3 \Delta[\mathrm{A}]}{2 \Delta t}
C −2Δ[A]3Δt\dfrac{-2 \Delta[\mathrm{A}]}{3 \Delta \mathrm{t}}
D Δ[A]Δt\dfrac{\Delta[\mathrm{A}]}{\Delta t}
Correct Answer
Option C
Solution
3 A→2 Br=−13Δ[A]Δt=+12Δ[B]Δt+Δ[B]Δt=−23Δ[A]Δt\begin{aligned} & 3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\ & \mathrm{r}=-\frac{1}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}=+\frac{1}{2} \frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}} \\ &+\frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}}=-\frac{2}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}} \end{aligned}
Q9
The correct options for the rate law that corresponds to overall first order reaction is
A Rate =k[A]0[ B]2=\mathrm{k}[\mathrm{A}]^0[\mathrm{~B}]^2
B Rate =k[A][B]=\mathrm{k}[\mathrm{A}][\mathrm{B}]
C Rate =k[A]1/2[ B]2=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^2
D Rate =k[A]−1/2[ B]3/2=\mathrm{k}[\mathrm{A}]^{-1 / 2}[\mathrm{~B}]^{3 / 2}
Correct Answer
Option D
Solution
r=k[A]−1/2[ B]3/2 order =−12+32=22=1\begin{aligned} & \mathrm{r}=\mathrm{k}[\mathrm{A}]^{-1 / 2}[\mathrm{~B}]^{3 / 2} \\ & \text{ order }=-\frac{1}{2}+\frac{3}{2} \\ & =\frac{2}{2} \\ & =1 \end{aligned}
Q10
For a certain reaction, the rate =k[A]2[ B]=\mathrm{k}[\mathrm{A}]^{2}[\mathrm{~B}], when the initial concentration of A is tripled keeping concentration of B\mathrm{B} constant, the initial rate would
A increase by a factor of six
B increase by a factor of nine
C increase by a factor of three
D decrease by a factor of nine
Correct Answer
Option B
Solution

Rate

=k[A]2[ B]=k[A]^{2}[\mathrm{~B}]

If

[A][A]

is tripled and

[B][B]

is kept constant.

r1=k[3A]2[B]r^{1}=k[3 A]^{2}[B]
r1=9k[A]2[B]r^{1}=9 k[A]^{2}[B]
r1=9rr^{1}=9 r

Increased by a factor of nine

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