Chemical Kinetics Test

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Chemical Kinetics Test - 45

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Question 1
1 / -0

$$T (in \,K)$$ $$\dfrac{1}{T} (in \, K^{-1})$$ $$\log_{10} K$$
$$769$$ $$1.3\times 10^{-3}$$ $$2.9$$
$$667$$ $$1.5\times 10^{-3}$$ $$1.1$$
From the following data: the activation energy for the reaction (cal/mol) $${ H }_{ 2 }+{ I }_{ 2 }\rightarrow 2HI$$

A
$$4\times { 10 }^{ 4 }$$

B
$$2\times { 10 }^{ 4 }$$

C
$$8\times { 10 }^{ 4 }$$

D
$$3\times { 10 }^{ 4 }$$

Solution

$$\displaystyle \log\frac{K_{2}}{K_{1}}=\frac{E_{act}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$$

$$\Rightarrow 2.9-1.1=\dfrac{E_{act}}{2}\times 0.2\times 10^{-3}$$

$$\Rightarrow E_{act}=\dfrac{2\times 1.8\times 10^{3}\times 10}{2}$$

$$=18000 \ cal$$

$$\approx 2\times 10^{4}cal $$
Question 2
1 / -0

If the door of a refrigerator is kept open in a dosed room then room:

A
heated

B
cooled

C
heated or cooled depending upon the initial temperature of the room

D
neither cooled nor heated

Solution

When the action of fridge is running, it takes air inside the fridge and cools it by removing heat from it and this heat come out from the back of fridge by a little fan. More heat comes outside the fridge. So if you even open the fridge door the cool air escape the fridge will heats up the room more ten cooling.
Question 3
1 / -0

In the case of a zero-order reaction, the ratio of time required for $$75$$% completion to $$50$$ % completion is:

A
$$ln 2$$

B
$$2$$

C
$$1.5$$

D
none

Solution

Its a Zero Order Reaction,

Time required for 75% completion
$${t}_{1} = \dfrac{100 -25}{2k}$$ = $$\dfrac{75}{2k}$$

Time required for 50% completion
$${t}_{2} = \dfrac{100 -50}{2k}$$ = $$\dfrac{50}{2k}$$

The Ratio of times = $$1.5$$
Question 4
1 / -0

Which integrated equation is correct for the following $$1^{st}$$ order reaction started with only $$A(g)$$ in a closed rigid vessel?
$$A(g)\rightarrow B(g) + C(g)+ D(g)$$

where, $$P_i= $$ initial pressure $$;\quad P_t= $$ total pressure at time $$t $$

A
$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_i }{P_t } \right]$$

B
$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_t }{P_i } \right]$$

C
$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 2P_i}{3P_i-P_t} \right]$$

D
$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 3P_i}{2P_i-3P_t} \right]$$

Solution

$$A\left( g \right) \longrightarrow B\left( g \right) +C\left( g \right) +D\left( g \right) $$
Initial $$Pi$$ $$0$$ $$0$$ $$0$$
at time $$t$$, $$Pi-p$$ $$p$$ $$p$$ $$p$$

$$\therefore$$ total pressure at $$t$$, $${ P }_{ t }={ P }_{ i }-p+p+p+p={ P }_{ i }+2p$$

Since the reaction is $${ 1 }^{ st }$$ order reaction.

$$\therefore$$ rate constant, $$K=\dfrac { 2.303 }{ t } log\left( \dfrac { { P }_{ i } }{ { P }_{ i }-p } \right) $$

$$\Rightarrow K=\dfrac { 2.303 }{ t } log\dfrac { { P }_{ i } }{ \left( { P }_{ i }-\dfrac { { P }_{ t }-{ P }_{ i } }{ 2 } \right) } $$

$$\Rightarrow K=\dfrac { 2.303 }{ t } log\left( \dfrac { 2{ P }_{ i } }{ { 3P }_{ i }-{ P }_{ t } } \right) $$ $$\therefore$$ Correct answer is C.
Question 5
1 / -0

The half-life for the reaction $$N_2O_5\to 2NO_2+\dfrac{1}{2}O_2$$ is $$2.4h$$ STP
Starting with $$10.8g$$ of $$N_2O_5$$ how much oxygen will be obtained after a peroid of $$9.6h$$?

A
1.5 L

B
3.36L

C
1.05 L

D
0.07L

Solution

Given $$t = 9.6h$$, $$n = \cfrac {9.6}{2.4} =4$$
Here n = numbers of half-life
$$N_2O_5 \rightarrow 2NO_2+1/2O_2$$ $$N_2=0.1×(1/2)^n$$
Moles of $$N_2O_5$$ left $$=\cfrac{0.1}{16}$$
Moles of $$N_2O_5$$ changed to product $$= (0.1−\cfrac{0.1}{16}$$)=$$\cfrac{1.5}{16}$$ mol
Moles of $$O_2$$ formed $$= \cfrac{1.5}{16}\times1/2=\cfrac{1.5}{32}$$
Volume of oxygen $$= 1.532×22.4=1.05\ L$$
Question 6
1 / -0

If the concentration of reactants is reduced by n times then the value of rate constant of the first order will?

A
Increase by n times

B
Decrease by factor of n

C
Not change

D
None of these

Solution

Solution:-(C) Not change
According to Arrhenius equation-
Rate constant $$\left( K \right) = A {e}^{-{{E}_{a}}/{RT}}$$
whereas,
$$T =$$ Temperature at which reaction is carried out.
$$A =$$ Arrhenius Parameter
$${E}_{a} =$$ Activation Energy of a reaction.
$$R =$$ Universal Gas Constant. .
For a particular reaction, rate constant depends on the temperature only and thus is independent of concentration of reactant.
Hence there will be no change in the value of rate constant w.r.t. the concentration of reactant.
Question 7
1 / -0

For a firt order reaction $$A\to P$$, the temperature (T) dependent rate constant (k) was found to follow the equation $$\log k = -(2000)\dfrac{1}{T}+6.0$$. The pre-exponential factor A and the activation energy $$E_a$$, respectively, are?

A
$$1.0\times 10^6s^{-1}$$ and $$9.2kJ\ mol^{-1}$$

B
$$6.0^{-1}$$ and $$16.6kJ\ mol^{-1}$$

C
$$1.0\times 10^6s^{-1}$$ and $$16.6kJ\ mol^{-1}$$

D
$$1.0\times 10^6s^{-1}$$ and $$38.3kJ\ mol^{-1}$$

Solution

$$log\space k = \dfrac{-2000}{T} + 6.0$$
Multiply by $$2.303$$
$$ln\space \dfrac{k}{1 \times {10}^{6}} = \dfrac{-4606}{T}$$
$$k = 1 \times {10}^{6}\space {e}^{\dfrac{-38.3 \times {10}^{3}}{RT}}$$
$$A = 1 \times {10}^{6}\space {s}^{-1}$$ and
$${E}_{a} = 38.3\space kJ\space {mol}^{-1}$$
Question 8
1 / -0

The time elapsed of a certain between 33% and 67% completion of a first order reaction is 30 minutes. What is the time needed for 25% completion?

A
$$150.5$$ minutes

B
$$12.5$$ minutes

C
$$180.5$$ minutes

D
$$165.5$$ minutes

Solution

$$30 = \dfrac{2.303}{k} log \dfrac{100}{33} - \dfrac{2.303}{k} log \dfrac{100}{67}$$

$$k = 0.0236\space {min}^{-1}$$

$$0.0236 = \dfrac{2.303}{t} log \dfrac{100}{75}$$

$$t = 12.5\space minutes$$
Question 9
1 / -0

For a reaction $$A \rightarrow B + C$$.. it was found that at the end of $$10$$ minutes from the start, the total optical rotation of the system was $$50^o$$ and when the reaction was completed a was $$100^o$$ Assuming that "$$A$$" a optically inactive. '$$B$$' dextro rotatory and '$$C$$' is laevo rotatory rate constant of this first order reaction is ?

A
$$0.069 \,min^{-1}$$

B
$$0.69 \,min^{-1}$$

C
$$2.303 \,min^{-1}$$

D
$$4.6 \,min^{-1}$$

Solution
Question 10
1 / -0

At $$373\ K$$, a gaseous reaction $$A\rightarrow 2B+C$$ is found to be first order. Starting with pure $$A$$, the total pressure at the end of $$10\ min$$. was $$176\ mm$$ and after a long time when $$A$$ was completely dissociated, it was $$270\ mm$$. The pressure of $$A$$ at the end of $$10$$ minutes was:

A
$$94\ mm$$

B
$$47\ mm$$

C
$$43\ mm$$

D
$$90\ mm$$

Solution

A single gaseous molecule produces 3 molecules. The pressure of the system is 270 mm.

$${p}_{a} = \dfrac{270}{3} = 90\space mm\space of\space Hg$$

If x is the decrease in the partial pressure of A at t = 10 min, then partial pressure of B and C is 2x and x

Total Pressure at this stage would be

$${p}_{b} + {p}_{c} + {p}_{a} = 176\space mm\space of\space Hg$$

90 -x + 2x + x = 176 mm of hg

$$x = 43 mm of Hg$$

So Option C is correct

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$$T (in \,K)$$	$$\dfrac{1}{T} (in \, K^{-1})$$	$$\log_{10} K$$
$$769$$	$$1.3\times 10^{-3}$$	$$2.9$$
$$667$$	$$1.5\times 10^{-3}$$	$$1.1$$

Chemical Kinetics Test - 45

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Chemical Kinetics Test - 45

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SHARING IS CARING

Question 1

$$4\times { 10 }^{ 4 }$$

$$2\times { 10 }^{ 4 }$$

$$8\times { 10 }^{ 4 }$$

$$3\times { 10 }^{ 4 }$$

Solution

Question 2

heated

cooled

heated or cooled depending upon the initial temperature of the room

neither cooled nor heated

Solution

Question 3

$$ln 2$$

$$2$$

$$1.5$$

none

Solution

Question 4

$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_i }{P_t } \right]$$

$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { P_t }{P_i } \right]$$

$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 2P_i}{3P_i-P_t} \right]$$

$$K=\dfrac { 2.303 }{ t } \log _{ 10 } \left[\dfrac { 3P_i}{2P_i-3P_t} \right]$$

Solution

Question 5

1.5 L

3.36L

1.05 L

0.07L

Solution

Question 6

Increase by n times

Decrease by factor of n

Not change

None of these

Solution

Question 7

$$1.0\times 10^6s^{-1}$$ and $$9.2kJ\ mol^{-1}$$

$$6.0^{-1}$$ and $$16.6kJ\ mol^{-1}$$

$$1.0\times 10^6s^{-1}$$ and $$16.6kJ\ mol^{-1}$$

$$1.0\times 10^6s^{-1}$$ and $$38.3kJ\ mol^{-1}$$

Solution

Question 8

$$150.5$$ minutes

$$12.5$$ minutes

$$180.5$$ minutes

$$165.5$$ minutes

Solution

Question 9

$$0.069 \,min^{-1}$$

$$0.69 \,min^{-1}$$

$$2.303 \,min^{-1}$$

$$4.6 \,min^{-1}$$

Solution

Question 10

$$94\ mm$$

$$47\ mm$$

$$43\ mm$$

$$90\ mm$$

Solution

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