Chemical Kinetics Test

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Chemical Kinetics Test - 57

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Question 1
1 / -0

$$A(g)\rightarrow B(g)+C(g)$$

$$\dfrac {-d[A]}{dt}=k[A]$$

At the start, the pressure is 100 mm and after 10 min, the pressure is 120 mm. Hence, rate constant $$(min^{-1})$$ is:

A
$$\dfrac {2.303}{10} log (\dfrac {120}{100})$$

B
$$\dfrac {2.303}{10} log (\dfrac {100}{20})$$

C
$$\dfrac {2.303}{10} log (\dfrac {100}{80})$$

D
$$\dfrac {2.303}{10} log( \dfrac {100}{120})$$

Solution

$$A(g)\rightarrow B(g) + C(g)$$

$$t=0$$ P 0 0
At time t P-x x x

Total pressure $$=P-x+x+x=120$$.
$$P+x=120$$

$$x=120-P=120-100=20.$$

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

$$\ k=\dfrac {2.303}{10} log \dfrac {100}{80}$$

Hence, option C is correct.
Question 2
1 / -0

For the $$1^{st}$$ order reaction: $$A_{(g)}\rightarrow 2B_{(g)}+C_{(s)}$$, the value of $$t_{\frac {1}{2}}=24\ mins$$. The reaction is carried out by taking a certain mass of $$A$$ enclosed in a vessel in which it exerts a pressure of $$400\ mm\ Hg$$. The pressure of the reaction mixture after the expiry of $$48\ mins$$ will be:

A
$$700\ mm$$

B
$$600\ mm$$

C
$$800\ mm$$

D
$$1000\ mm$$

Solution
Question 3
1 / -0

The rate constant $$(k)$$ of a first-order reaction is $$0.0693\ min^{-1}$$. If we start with $$20\ mol\ L^{-1}$$, then it is reduced to $$2.5\ mol\ L^{-1}$$ in:

A
$$10\ mins$$

B
$$20\ mins$$

C
$$30\ mins$$

D
$$40\ mins$$

Solution
Question 4
1 / -0

In the first order reaction the concentration of reactant decreases from $$1.0\ M$$ to $$0.25\ M$$ in $$20$$ minutes. The value of specific rate is:

A
$$69.32$$

B
$$6.932$$

C
$$0.6932$$

D
$$0.06932$$

Solution

$$\frac {2.303}{20} log \frac {2}{0.5}=0.06932$$.
Question 5
1 / -0

Calculate the half-life of the first-order reaction, $$C_2H_4O(g)\rightarrow CH_4(g)+CO(g)$$, if the initial pressure of $$C_2H_4O(g)$$ is 80 mm and the total pressure at the end of 20 minutes is 120 mm.

A
40 min

B
120 min

C
20 min

D
80 min

Solution

For first-order reaction,
$$k=\dfrac {2.303}{t} log \dfrac {a}{a-x}$$

$$=\dfrac {2.303}{20}\times log\dfrac {80}{40}$$

$$=\dfrac {2.303}{20}\times 0.3010$$

For half-life period,

$$t_{\frac {1}{2}}=\dfrac {0.693}{k}=\dfrac {0.693\times 20}{2.303\times 3010}=20\ min$$

Hence, option C is correct.
Question 6
1 / -0

A reaction that is of the first order with respect to reactant A has a rate constant 6 min$$^{-1}$$. If we start with [A]=0.5 mol L$$^{-1}$$, when would [A] reach the value of 0.05 mol L$$^{-1}$$?

A
0.384 min

B
0.15 min

C
3 min

D
3.84 min

Solution

We know that for first order kinetics:

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

or $$t=\dfrac {2.303}{6} log \dfrac {0.5}{0.05}=\dfrac {2.303}{6}=0.384\ min$$

Hence, option A is correct.
Question 7
1 / -0

For the first order reaction $$A_{(g)}\rightarrow 2B_{(g)}+C_{(g)}$$, the initial pressure is $$P_A=90 mm$$ Hg, the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is:

A
$$1.15\times 10^{-3} sec^{-1}$$

B
$$2.3\times 10^{-3} sec^{-1}$$

C
$$3.45\times 10^{-3} sec^{-1}$$

D
$$6\times 10^{-3} sec^{-1}$$

Solution

$$A\rightarrow 2B+C$$
P 0 0
P-x 2x x
At equilibrium,
$$180=P-x+2x+x$$

$$180=90+2x$$

$$2x=90, x=45$$

$$k=\dfrac {2.303}{t} log \dfrac {P}{P-x}$$

$$=\dfrac {2.303}{10} log\dfrac {90}{90-45}=\dfrac {2.303}{10} log 2=\dfrac {0.6932}{10}$$

$$=0.6932=\dfrac {0.06932}{60}=1.1555\times 10^{-3} sec^{-1}$$.
Question 8
1 / -0

A certain zero order reaction has $$k=0.025\ M sec^{-1}$$ for the disappearance of A. What will be the concentration of A after 15 seconds if initial concentration is 0.5 M?

A
$$0.5 \ M$$

B
$$0.32 \ M$$

C
$$0.125 \ M$$

D
$$0.06 \ M$$

Solution

$$x=kt=0.025\times 15=0.375\ M$$
Remaining concentration of $$A= 0.5-0.375$$
$$=0.125\ M$$
Question 9
1 / -0

In the first order reaction, the concentration of the reactant is reduced to 25% in one hour.The half-life period of the reaction is:

A
2 hr

B
4 hr

C
1/2 hr

D
1/4 hr

Solution

$$k=\frac {2.303}{t} log \frac {a}{a-x}$$
$$k=\frac {2.303}{1} log \frac {100}{25}=\frac {2.303}{1} log 4=2.303\times 2\times 0.3010$$
$$t_{\frac {1}{2}}=\frac {0.693}{k}\therefore t_{\frac {1}{2}}=0.5 hr$$.
Question 10
1 / -0

For a 1$$^{st}$$ order reaction (gaseous) (constant V, T) $$a A \rightarrow (b - 1) B + 1 C$$ (with b > a) the pressure of the system rose by $$50 \left ( \frac{b}{a} - 1 \right )$$% in a time of 10 min. The half life of the reaction is therefore:

A
10 min

B
20 min

C
30 min

D
40 min

Solution

$$aA \rightarrow$$ (b -1) B + C
$$t = 0;$$ $$P_0$$ 0 0
$$t= t;$$ $$P_0 - x$$ $$\frac{(b-1)x}{a}$$ $$\frac{x}{a}$$

Total pressure $$=P_0 - x + \frac{b}{a} x = P_0 + \left ( \frac{b}{a} - 1 \right ) \frac{1}{2} P_0$$

$$P_0 + \left ( \frac{b}{a} - 1 \right ) x = P_0 + \left ( \frac{b}{a} - 1 \right ) \frac{P_0}{2}$$

$$x = \frac{P_0}{2}$$

$$P_A = P_0 - x = P_0 - \frac{P_0}{2} = \frac{P_0}{2}$$

$$P_A$$ reduces to half in 10 min.

So, $$t_{1/2} = 10\ min.$$

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Chemical Kinetics Test - 57

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Chemical Kinetics Test - 57

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SHARING IS CARING

Question 1

$$\dfrac {2.303}{10} log (\dfrac {120}{100})$$

$$\dfrac {2.303}{10} log (\dfrac {100}{20})$$

$$\dfrac {2.303}{10} log (\dfrac {100}{80})$$

$$\dfrac {2.303}{10} log( \dfrac {100}{120})$$

Solution

Question 2

$$700\ mm$$

$$600\ mm$$

$$800\ mm$$

$$1000\ mm$$

Solution

Question 3

$$10\ mins$$

$$20\ mins$$

$$30\ mins$$

$$40\ mins$$

Solution

Question 4

$$69.32$$

$$6.932$$

$$0.6932$$

$$0.06932$$

Solution

Question 5

40 min

120 min

20 min

80 min

Solution

Question 6

0.384 min

0.15 min

3 min

3.84 min

Solution

Question 7

$$1.15\times 10^{-3} sec^{-1}$$

$$2.3\times 10^{-3} sec^{-1}$$

$$3.45\times 10^{-3} sec^{-1}$$

$$6\times 10^{-3} sec^{-1}$$

Solution

Question 8

$$0.5 \ M$$

$$0.32 \ M$$

$$0.125 \ M$$

$$0.06 \ M$$

Solution

Question 9

2 hr

4 hr

1/2 hr

1/4 hr

Solution

Question 10

10 min

20 min

30 min

40 min

Solution

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