Chemical Kinetics Test

Result

Chemical Kinetics Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

First order reaction completes $$20\%$$ in $$5$$ min. How much time it will take for $$60\%$$ completion?

A
$$26.5$$ min

B
$$20.5$$ min

C
$$19.5$$ min

D
$$18$$ min

Solution

In 5 minutes, the reaction is 20% complete.
$$\displaystyle [A]_0=100$$ and $$\displaystyle [A]=100-20=80$$

$$\displaystyle k= \dfrac {2.303}{t} log_{10} \dfrac {[A]_0}{[A]} $$

$$\displaystyle k= \dfrac {2.303}{5 \ min} log_{10} \dfrac { 100}{80} $$

$$\displaystyle k=0.0446 \ min^{-1}$$

Now, the reaction is 60% complete.

$$\displaystyle [A]_0=100$$ and $$\displaystyle [A]=100-60=40$$

$$\displaystyle t= \dfrac {2.303}{k} log_{10} \dfrac {[A]_0}{[A]} $$

$$\displaystyle t= \dfrac {2.303}{0.0446 \ min^{-1}} log_{10} \dfrac { 100}{40} $$

$$\displaystyle t=20.5 \ min$$
Question 2
1 / -0

A first order reaction takes $$40$$ min for $$30$$% decomposition. What will be $$t_{1/2}$$?

A
$$77.7\ min.$$

B
$$52.5\ min.$$

C
$$46.2\ min.$$

D
$$22.7\ min.$$

Solution

$$30$$% decomposition means $$X = 30$$% of $$R_{0}$$ or $$R = R_{0} - 0.3R_{0} = 0.7R_{0}$$

For first order, $$k = \dfrac {2.303}{t}\log \dfrac {[R_{0}]}{[R]}$$

$$= \dfrac {2.303}{40}\log \dfrac {10}{7} min^{-1} = 8.918\times 10^{-3} min.^{-1}$$

$$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{8.918\times 10^{-3} min.^{-1}} = 77.7\ min/$$
Question 3
1 / -0

Cyclopropane rearranges to form propene:

$$\triangle \rightarrow CH_{2} - CH = CH_{2}$$

This follows first-order kinetics. The rate constant is $$2.714\times 10^{-3} sec^{-1}$$. The initial concentration of cyclopropane is $$0.29\ M$$. What will be the concentration of cyclopropane after $$100\ sec$$?

A
$$0.035\ M$$

B
$$0.22\ M$$

C
$$0.145\ M$$

D
$$0.0018\ M$$

Solution

$$k = \dfrac {2.303}{t} \log \dfrac {a}{(a - x)}$$

$$(a - x)$$ is the concentration left after $$100\ sec$$.

$$2.7\times 10^{-3} = \dfrac {2.303}{100} \log \dfrac {0.29}{(a - x)}$$

$$\Rightarrow \dfrac {0.27}{2.303} = \log \dfrac {0.29}{(a - x)} \Rightarrow 0.117 = \log \dfrac {0.29}{(a -x)}$$

$$\Rightarrow (a - x) = 0.22\ M$$.

Hence, the correct answer is option $$\text{B}$$.
Question 4
1 / -0

The reaction $$2X \rightarrow Y + Z$$ would be zero order reaction when:

A
rate remains unchanged at any concentration of $$Y$$ and $$Z$$

B
rate of reaction doubles if concentration of $$Y$$ is doubled

C
rate of reaction remains same at any concentration of $$X$$

D
rate of reaction of directly proportional to square of concentration of $$X$$

Solution

By definition, the zero-order reaction is that type of reaction in which the rate of reaction is independent of the concentration of reactants or product. So, the rate of reaction remains the same at any concentration of X.
Question 5
1 / -0

The decomposition of a substance follows first order kinetics. If its concentration is reduced to $$1/8$$ of its initial value in $$12$$ minutes, the rate constant of the decomposition system is ?

A
$$\left (\dfrac {2.303}{12}\log \dfrac {1}{8}\right ) min.^{-1}$$

B
$$\left (\dfrac {2.303}{12}\log 8\right ) min.^{-1}$$

C
$$\left (\dfrac {0.693}{12}\right ) min.^{-1}$$

D
$$\left (\dfrac {1}{12}\log 8\right )min.^{-1}$$

Solution

$$k = \dfrac {2.303}{t} \log\left( \dfrac {a}{a - x}\right)$$ (for first order)

$$k = \dfrac {2.303}{12}\log\left( \dfrac {1}{1/8}\right)= \left (\dfrac {2.303}{12}\log 8\right ) min.^{-1}$$.
Question 6
1 / -0

For the first order reaction,
$$2H_2O_2\rightarrow 2H_2\,+\,O_2 \,\log k\,=\,14.34\,-\,1.25\,\times\,10^4\,K/T$$. The energy of activation for the above reaction if its half life period is 256 min.

A
400 KJ / mole

B
230 KJ / mole

C
480 KJ / mole

D
239 KJ / mole

Solution

Given order of the reaction= First order
$$t_{1/2}=256min$$
$$K= \cfrac {0.693}{t_{1/2}}=\cfrac {0.693}{256\times 60}=8.67 \times 10^{-8}$$
Given, $$\log K= 14.34- \cfrac {1.25 \times 10^4}{T}K \longrightarrow (1)$$
By Arrhenius equation,
$$\log K= \log A- \cfrac {E_a}{2.303RT} \longrightarrow (2)$$
(1) & (2) $$\longrightarrow$$
$$\Rightarrow \cfrac {E_a}{2.303RT}= \cfrac {1.25 \times 10^4}{T}$$
$$\Rightarrow E_a= 1.25 \times 10^4\times 2.303\times 8.314$$
$$= 2.39 \times 10^5$$
$$=239 kJ/mole$$
Question 7
1 / -0

At low pressure, the fraction of the surface covered follows:

A
zero-order kinetics

B
first order kinetics

C
second order kinetics

D
fractional order kinetics

Solution

At low pressure $$(P_A \rightarrow 0)$$, $$\theta_A$$ is very small and proportional to the pressure.
The rate becomes first order rate (low P) = $$k_2K_AP_A$$
If a substance is adsorbed on surfaces.
It is the first order reaction.
Question 8
1 / -0

The following data were obtained during the first order thermal decomposition of $$SO_2Cl_2$$ at constant volume
$$SO_2Cl_{2(g)} \rightarrow SO_{2(g)}\,+\,Cl_{2(g)}$$
Expt Time / $$S^{-1}$$ Total pressure/ atm
1 0 0.5
2 100 0.6
when the total pressure is 0.65 atm then the rate constant of reaction is ..............
if a = pi, and a - x = 2pi - Pt

A
2.0 x $$10^{-3}\,S^{-1}$$

B
2.12 x $$10^{-3}\,S^{-1}$$

C
2.23 x $$10^{-3}\,S^{-1}$$

D
2.34 x $$10^{-3}\,S^{-1}$$

Solution

First order thermal decomposition $$SO_2 \, Cl_2 (g) \rightarrow SO_2 (g) + Cl_2 (g)$$
$$\begin{matrix} at\, t=0 & P_{ 0 } & 0 & 0 \\ at\, t=100 & P_{ 0 }-P & P & P \end{matrix}$$
$$P_0 = 0.5\ atm$$
$$P_0 + P = 0.6$$
$$P = 0.6 - 0.5 = 0.1$$
$$k = \dfrac{1}{t} \, \ln \, \dfrac{P_0}{P_0 - P}$$
$$k = \dfrac{1}{100} \, \ln \, \dfrac{0.5}{0.4}$$
$$k = \dfrac{1}{100} \, \ln \, 1.25$$
$$= 0.223 \times 10^{-2} = 2.23 \times 10^{-3}$$
option $$C$$ is correct option
Question 9
1 / -0

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 $$3.0\ g$$?

A
$$34.07\ s$$

B
$$7.57\ s$$

C
$$10.10\ s$$

D
$$15\ s$$

Solution

$$t = \dfrac {2.303}{k}\log \left(\dfrac {a}{(a - x)}\right)$$

or $$t = \dfrac {2.303}{15\times 10^{-3}} \log\left( \dfrac {5}{3}\right) = 34.00\ s$$
Question 10
1 / -0

The half-life of the reaction $$X\rightarrow Y$$, following first-order kinetics, when the initial concentration of $$X$$ is $$0.01$$ mol L$$^{-1}$$ and the initial rate is $$0.00352$$ mol L$$^{-1}$$ min$$^{-1}$$ will be :

A
$$19.60\ min$$

B
$$1.969\ min$$

C
$$7.75\ min$$

D
$$77.5\ min$$

Solution

For a first order reaction,

$$\dfrac {dx}{dt} = k[X]$$

$$0.00352 = k\times 0.01$$

$$\Rightarrow k = 0.352$$

$$t_{1/2} = \dfrac {0.693}{k} = \dfrac {0.693}{0.352} = 1.969\ min$$

Hence, the correct answer is option $$\text{B}$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Expt	Time / $$S^{-1}$$	Total pressure/ atm
1	0	0.5
2	100	0.6

Chemical Kinetics Test - 39

Result

Chemical Kinetics Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$26.5$$ min

$$20.5$$ min

$$19.5$$ min

$$18$$ min

Solution

Question 2

$$77.7\ min.$$

$$52.5\ min.$$

$$46.2\ min.$$

$$22.7\ min.$$

Solution

Question 3

$$0.035\ M$$

$$0.22\ M$$

$$0.145\ M$$

$$0.0018\ M$$

Solution

Question 4

rate remains unchanged at any concentration of $$Y$$ and $$Z$$

rate of reaction doubles if concentration of $$Y$$ is doubled

rate of reaction remains same at any concentration of $$X$$

rate of reaction of directly proportional to square of concentration of $$X$$

Solution

Question 5

$$\left (\dfrac {2.303}{12}\log \dfrac {1}{8}\right ) min.^{-1}$$

$$\left (\dfrac {2.303}{12}\log 8\right ) min.^{-1}$$

$$\left (\dfrac {0.693}{12}\right ) min.^{-1}$$

$$\left (\dfrac {1}{12}\log 8\right )min.^{-1}$$

Solution

Question 6

400 KJ / mole

230 KJ / mole

480 KJ / mole

239 KJ / mole

Solution

Question 7

zero-order kinetics

first order kinetics

second order kinetics

fractional order kinetics

Solution

Question 8

2.0 x $$10^{-3}\,S^{-1}$$

2.12 x $$10^{-3}\,S^{-1}$$

2.23 x $$10^{-3}\,S^{-1}$$

2.34 x $$10^{-3}\,S^{-1}$$

Solution

Question 9

$$34.07\ s$$

$$7.57\ s$$

$$10.10\ s$$

$$15\ s$$

Solution

Question 10

$$19.60\ min$$

$$1.969\ min$$

$$7.75\ min$$

$$77.5\ min$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.