Chemical Kinetics Test

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Chemical Kinetics Test - 61

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Question 1
1 / -0

The reaction $$A(g)\longrightarrow B(g)+2C(g)$$ is a first order reaction with rate constant $$2.772\times { 10 }^{ -3 }s^{ -1 }$$. Starting with $$0.1$$ mole of $$A$$ in $$2$$ litre vessel, find the concentration of $$A$$ after $$250sec$$ when the reaction is allowed to take place at constant pressure at $$300K$$? (Given ln2= 0.693)

A
$$0.0125M$$

B
$$0.025M$$

C
$$0.05M$$

D
none of these

Solution

Half life of reaction is $$t_{1/2} = 0.693/k = 250$$ sec.
So after 250 sec concentration become half and remaing concentration is $$0.05/2 = 0.025M$$.
Question 2
1 / -0

For a given reaction of the first order, it takes 15 minutes for the concentration to drop from 0.8 M to 0.4 M. The time required for the concentration to drop from 0.1 M to 0.025 M will be :

A
15 minutes

B
60 minutes

C
30 minutes

D
7.5 minutes

Solution

The reactant concentration drop from $$0.8\ M$$ to $$0.4\ M$$, i.e., $$50\%$$ takes place in 15 minutes.

$$K=\dfrac {2.303}{15} log \dfrac {0.8}{0.4}=\dfrac {0.693}{15}=0.0462 \ min^{-1}$$

Also, $$t=\dfrac {2.303}{K}log \dfrac {0.1}{0.025}$$ $$=\dfrac {2.303}{0.0462} log \dfrac {0.1}{0.025}=30\ min$$
Question 3
1 / -0

A reaction occurs in parallel paths. For each path having energy of activation as $$E,\ 2E,\ 3E,\ ... nE$$ and rate constant $$K,\ 2K,\ 3K,\ .... nK$$ respectively. If $$E_{AV}=3E$$, then find out the value of $$n$$.

A
$$4$$

B
$$5$$

C
$$6$$

D
None of these

Solution

The average energy is given by the expression $$ \displaystyle 3E=\frac {E+2E+3E+...+nE}{n}=\frac {n(n+1)}{2n}E$$

Hence, $$ \displaystyle\frac {n+1}{2}=3$$

$$6=n+1$$

$$n=5$$
Question 4
1 / -0

The decomposition of $$Cl_2O_7$$ at 400 K in gaseous phase to $$Cl_2$$ and $$O_2$$ is 1st order reaction. After 55 sec at 400 K, the pressure of reaction mixture increases from 0.62 to 1.88 atm. Calculate the rate constant of the reaction. Also, calculate the pressure of reaction mixture after 100 seconds:

A
$$1.58\times 10^{-2}, 2.33 atm$$

B
$$1.58\times 10^{-3}, 2.33 atm$$

C
$$15.8\times 10^{-2}, 23.3 atm$$

D
None of these

Solution

$$\displaystyle Cl_2O_7 (g) \rightarrow Cl_2(g) + 3.5 O_2 (g) $$

Initial pressure of $$\displaystyle Cl_2O_7 = 0.62$$ atm
Equilibrium pressure of $$\displaystyle Cl_2O_7 = 0.62 - x$$ atm
Equilibrium pressure of $$\displaystyle Cl_2 = x$$ atm
Equilibrium pressure of $$\displaystyle O_2 = 3.5 x$$ atm.

Total equilibrium pressure $$\displaystyle = 0.62 - x + x + 3.5 x = 0.62 + 3.5 x = 1.88$$ atm
$$\displaystyle 3.5 x = 1.26$$
$$\displaystyle x = 0.36$$ atm

Hence the equilibrium pressure of $$\displaystyle Cl_2O_7 = 0.62 -x = 0.62 - 0.36 = 0.26$$ atm.

The rate constant of the reaction $$\displaystyle k = \dfrac {2.303}{t}log \dfrac {a}{a-x}$$

$$\displaystyle k = \dfrac {2.303}{55 \: sec} log \dfrac {0.62}{0.26} = 1.58 \times 10^{-2} /sec$$

After 100 sec,
$$\displaystyle 1.58 \times 10^{-2} = \dfrac {2.303}{100} log \dfrac {0.62}{a-x} $$

$$\displaystyle 0.6861 = log \dfrac {0.62}{a-x}$$

$$\displaystyle 4.854 = \dfrac {0.62}{a-x}$$

$$\displaystyle a-x = 0.1277$$

$$\displaystyle x = a - 0.1277 = 0.62 - 0.1277 = 0.4923$$

Total equilibrium pressure $$\displaystyle = 0.62 + 3.5 x = 0.62 + 3.5 (0.4923) = 2.33 $$ atm
Question 5
1 / -0

A compound $$A$$ dissociate by two parallel first order paths at certain temperature
$$A(g)\xrightarrow [ ]{ { k }_{ 1 }({ min }^{ -1 }) } 2B(g)$$ $${ k }_{ 1 }=6.93\times { 10 }^{ -3 }min^{ -1 }$$
$$A(g)\xrightarrow [ ]{ { k }_{ 2 }({ min }^{ -1 }) } C(g)$$ $${ k }_{ 2 }=6.93\times { 10 }^{ -3 }min^{ -1 }$$
The reaction started with $$1$$ mole of pure '$$A$$' in $$1$$ litre closed container with initial pressure $$2$$ atm. What is the pressure (in atm) developed in container after $$50$$ minutes from start of experiment?

A
$$1.25$$

B
$$0.75$$

C
$$1.50$$

D
$$2.50$$

Solution

For both reactions, half life is 100 min so after 50 min no. of moles of all reagents in solution is 1.25 mole. So the pressure of the container is $$2 \times 1.25/1 = 2.50$$ atm.
Question 6
1 / -0

The reaction $$cis-X\overset { { k }_{ f } }{ \underset { { k }_{ b } }{ \rightleftharpoons } } tran-X$$ is first order in both directions. At $${ 25 }^{ o }C$$, the equilibrium constant is $$0.10$$ and the rate constant $$ { k }_{ f }=3\times { 10 }^{ -4 }s^{ -1 }$$. In an experiment starting with the pure $$cis-$$form, how long would it take for half of the equilibrium amount of the $$trans-$$ isomer to be formed?

A
$$150$$sec

B
$$200$$sec

C
$$155$$sec

D
$$210$$sec

Solution
Question 7
1 / -0

A substance reacts according to 1$$^{st}$$ order kinetic and rate constant for the reaction is $$1\times 10^{-2} sec^{-1}$$. If the initial concentration is 1 M. Rate of the reaction after 1 minute is:

A
$$5.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

B
$$5.55\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

C
$$7.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

D
None of these

Solution

$$K=\dfrac {2.303}{t} log \dfrac {a}{(a-X)}$$

$$\therefore 10^{-2}=\dfrac {2.303}{1\times 60} log \dfrac {1}{(1-X)} (\because t=1 minute = 60 sec)$$

$$\therefore (1-X)=0.549$$

Therefore, rate after 1 minute $$=K[1-X]^1$$

$$=10^{-2}\times (0.549)=5.49\times 10^{-3} mol\ L^{-1} sec^{-1}$$

Hence, the correct option is $$A$$
Question 8
1 / -0

Mathematical expression for $$t_{1/4}$$ i.e., when $$(1/4)^{th}$$ reaction is over following first order kinetics can be given by:

A
$$t_{1/4}=\frac {2.303}{K}log 4$$

B
$$t_{1/4}=\frac {2.303}{K}log 1/4$$

C
$$t_{1/4}=\frac {2.303}{K}log 2$$

D
$$t_{1/4}=\frac {2.303}{K}log \frac {4}{3}$$

Solution

As we know, for a first order reaction,
$$t=\dfrac {2.303}{K}log \dfrac {a}{(a-x)}$$

If $$t=t_{1/4}; x=\dfrac a4$$

$$\therefore t_{1/4}=\dfrac {2.303}{K} log \dfrac {a}{a-a/4}$$ $$=\dfrac {2.303}{K} log\dfrac {4}{3}$$
Question 9
1 / -0

Catalytic decomposition of nitrous oxide by gold at $$900^oC$$ at an initial pressure of $$200$$ mm, was $$50$$% in $$53$$ minutes and $$73$$% in $$100$$ minutes.
Velocity constant of the reaction is:
[Note: assume decomposition as the first order.]

A
$$1.308\times 10^{-2}$$

B
$$2.317\times 10^{-2}$$

C
$$3.208\times 10^{-3}$$

D
none of these

Solution

Taking decomposition as first order:

$$\because K=\dfrac {2.303}{t}log \dfrac {a}{(a-X)}$$

Case $$I$$: $$a\propto\ 200\ mm$$;

$$X\propto 200\times (50/100)\ mm$$ and $$t_{1/2}=53\ minutes$$

$$\therefore K_1=\dfrac {2.303}{53} log \dfrac {200}{200-100}$$

$$\quad \quad=1.308\times 10^{-2}\ min^{-1}$$

Case $$II$$: $$a\propto 200\ mm; X\propto 200\times (73/ 100)\propto 146 mm$$

$$\therefore K_2=\dfrac {2.303}{100}log \dfrac {200}{200-146}$$

$$\quad \quad =1.309\times 10^{-2} min^{-1}$$

$$K=\dfrac {K_1+K_2}{2}=\dfrac {(1.308+1.309)\times 10^{-2}}{2}=1.308\times 10^{-2}$$

Hence, the correct option is $$A$$
Question 10
1 / -0

In the thermal decomposition of $$N_2O$$ at 830 K, the time required to decompose half of the reactant was 263 sec at the initial pressure 290 mm. It takes 212 sec to decompose half of the reactant if initial pressure was 360 mm. What is the order of the reaction? Also calculate $$t_{1/2}$$ for $$N_2O$$ decomposition if initial pressure of $$N_2O$$ is 1 atm.

A
$$O.R=2 $$, Half life $$=100.2 sec $$

B
$$O.R=1 $$, Half life $$=10.02 sec $$

C
$$O.R=4 $$, Half life $$=1.002 sec $$

D
None of these

Solution

$$\displaystyle \dfrac {t_{1/2,1}}{t_{1/2, 2}} = \dfrac {263}{212} = 1.24$$

$$\displaystyle \dfrac {P_2}{P_1} = \dfrac {360}{290} = 1.24$$

Thus $$\displaystyle \dfrac {t_{1/2, 1}}{t_{1/2, 2}} = \dfrac {P_2}{P_1}$$

$$\displaystyle t_{1/2} \propto \dfrac {1}{P_i}$$

Thus, the half life period is inversely proportional to the initial pressure. This is characteristic of second order reaction.

$$\displaystyle \dfrac {t_{1/2, }}{263} = \dfrac {290}{760}$$ (because 1 atm = 760 mm Hg)

$$\displaystyle t_{1/2} = 100.2 sec$$

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 61

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Chemical Kinetics Test - 61

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SHARING IS CARING

Question 1

$$0.0125M$$

$$0.025M$$

$$0.05M$$

none of these

Solution

Question 2

15 minutes

60 minutes

30 minutes

7.5 minutes

Solution

Question 3

$$4$$

$$5$$

$$6$$

None of these

Solution

Question 4

$$1.58\times 10^{-2}, 2.33 atm$$

$$1.58\times 10^{-3}, 2.33 atm$$

$$15.8\times 10^{-2}, 23.3 atm$$

None of these

Solution

Question 5

$$1.25$$

$$0.75$$

$$1.50$$

$$2.50$$

Solution

Question 6

$$150$$sec

$$200$$sec

$$155$$sec

$$210$$sec

Solution

Question 7

$$5.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

$$5.55\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

$$7.49\times 10^{-3} mol\ litre^{-1} sec^{-1}$$

None of these

Solution

Question 8

$$t_{1/4}=\frac {2.303}{K}log 4$$

$$t_{1/4}=\frac {2.303}{K}log 1/4$$

$$t_{1/4}=\frac {2.303}{K}log 2$$

$$t_{1/4}=\frac {2.303}{K}log \frac {4}{3}$$

Solution

Question 9

$$1.308\times 10^{-2}$$

$$2.317\times 10^{-2}$$

$$3.208\times 10^{-3}$$

none of these

Solution

Question 10

$$O.R=2 $$, Half life $$=100.2 sec $$

$$O.R=1 $$, Half life $$=10.02 sec $$

$$O.R=4 $$, Half life $$=1.002 sec $$

None of these

Solution

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