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CHEMISTRY(KCET)

285442 $100 {cm}^{3}$ of $1 M {CH}_{3} COOH$ was mixed with 100 ${cm}^{3}$ of $2 M {CH}_{3} OH$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

1 4 times

2 0.25 times

3 2 times

4 0.5 times.

Explanation:

(b) ${CH}_{3} COOH + {CH}_{3} OH \to$
${CH}_{3} {COOCH}_{3} + H_{2} O$
$r_{1} = k [{CH}_{3} COOH] [{CH}_{3} OH] = k (1) (2) = 2 k$
When solution is diluted with equal volume of water, concentration is reduced to half.
$∴ r_{2} = k (\frac{1}{2}) (\frac{2}{2}) = \frac{k}{2}$
$\frac{r_{1}}{r_{2}} = \frac{2 k}{k / 2} = 4 \Rightarrow r_{2} = \frac{r_{1}}{4} = 0.25 r_{1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285443 In a first order reaction, the concentration of the reactant is reduced to $12.5 %$ in one hour. When was it half completed?

1 20 min

2 15 min

3 3 hr

4 30 min

Explanation:

(a) For a first order reaction,
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{60} \log \frac{100}{12.5}$
$= 0.0346 \min^{- 1}$
Now $t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0346} = 20 \min$

Karnataka CET 2015

CHEMISTRY(KCET)

285444 Half-life period of a first order reaction is 10 min . Starting with initial concentration 12 M , the rate after 20 min is

1

0.693 \times 3 M \min^{- 1}

2

0.0693 \times 4 \min^{- 1}

3

0.0693 M \min^{- 1}

4

0.0693 \times 3 M \min^{- 1}

Explanation:

(d) $12 M \overset{t_{1 / 2}}{\to} 6 M \overset{t_{1 / 2}}{\to} 3 M$
Initial conc.
$T_{1 / 2} = 10 \min$
$k = \frac{0.693}{10} = 0.0693 \min^{- 1}$
As $t_{1 / 2}$ is 10 min , after 20 minutes the concentration will be 3 M .
Hence, Rate $= 0.0693 \times 3 M \min^{- 1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285445 $A (g) \overset{Δ}{\to} P (g) + Q (g) + R (g)$ , follows first order kinetics with a half-life of 69.3 s at $500^{\circ} C$ . Starting from the gas 'A' enclosed in a container at $500^{\circ} C$ and at a pressure of 0.4 atm , the total pressure of the system after 230 s will be

1 1.15 atm

2 1.32 atm

3 1.22 atm

4 1.12 atm

Explanation:

(d) $T_{1 / 2} = 69.3 s, t = 230 s, a = 0.4 atm$ ,
For first order reaction,
$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{69.3} = 0.01 s^{- 1}$
$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{230} \log \frac{0.4}{0.4 - x}$
$0.01 = \frac{2.303}{230} \log \frac{0.4}{0.4 - x} = 0.01 \log \frac{0.4}{0.4 - x}$
- $\frac{0.01}{0.01} = \log \frac{0.4}{0.4 - x} \Rightarrow 1 = \log \frac{0.4}{0.4 - x}$
Antilog of $1 = \frac{0.4}{0.4 - x}$
$\Rightarrow 10 = \frac{0.4}{0.4 - x} \Rightarrow x = 0.36$
For the given reaction,
$\begin{array}{lcccc} A (g) \overset{Δ}{\to} & P (g) + Q (g) + & R (g) \\ Initial & 0.4 & 0 & 0 & 0 \\ Final & 0.4 - 0.36 & 0.36 & 0.36 & 0.36 \end{array}$
Total pressure $= (0.4 - 0.36) + (3 \times 0.36)$ $= 1.12 atm$

Karnataka CET 2014

CHEMISTRY(KCET)

285442 $100 {cm}^{3}$ of $1 M {CH}_{3} COOH$ was mixed with 100 ${cm}^{3}$ of $2 M {CH}_{3} OH$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

1 4 times

2 0.25 times

3 2 times

4 0.5 times.

Explanation:

(b) ${CH}_{3} COOH + {CH}_{3} OH \to$
${CH}_{3} {COOCH}_{3} + H_{2} O$
$r_{1} = k [{CH}_{3} COOH] [{CH}_{3} OH] = k (1) (2) = 2 k$
When solution is diluted with equal volume of water, concentration is reduced to half.
$∴ r_{2} = k (\frac{1}{2}) (\frac{2}{2}) = \frac{k}{2}$
$\frac{r_{1}}{r_{2}} = \frac{2 k}{k / 2} = 4 \Rightarrow r_{2} = \frac{r_{1}}{4} = 0.25 r_{1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285443 In a first order reaction, the concentration of the reactant is reduced to $12.5 %$ in one hour. When was it half completed?

1 20 min

2 15 min

3 3 hr

4 30 min

Explanation:

(a) For a first order reaction,
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{60} \log \frac{100}{12.5}$
$= 0.0346 \min^{- 1}$
Now $t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0346} = 20 \min$

Karnataka CET 2015

CHEMISTRY(KCET)

285444 Half-life period of a first order reaction is 10 min . Starting with initial concentration 12 M , the rate after 20 min is

1

0.693 \times 3 M \min^{- 1}

2

0.0693 \times 4 \min^{- 1}

3

0.0693 M \min^{- 1}

4

0.0693 \times 3 M \min^{- 1}

Explanation:

(d) $12 M \overset{t_{1 / 2}}{\to} 6 M \overset{t_{1 / 2}}{\to} 3 M$
Initial conc.
$T_{1 / 2} = 10 \min$
$k = \frac{0.693}{10} = 0.0693 \min^{- 1}$
As $t_{1 / 2}$ is 10 min , after 20 minutes the concentration will be 3 M .
Hence, Rate $= 0.0693 \times 3 M \min^{- 1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285445 $A (g) \overset{Δ}{\to} P (g) + Q (g) + R (g)$ , follows first order kinetics with a half-life of 69.3 s at $500^{\circ} C$ . Starting from the gas 'A' enclosed in a container at $500^{\circ} C$ and at a pressure of 0.4 atm , the total pressure of the system after 230 s will be

1 1.15 atm

2 1.32 atm

3 1.22 atm

4 1.12 atm

Explanation:

(d) $T_{1 / 2} = 69.3 s, t = 230 s, a = 0.4 atm$ ,
For first order reaction,
$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{69.3} = 0.01 s^{- 1}$
$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{230} \log \frac{0.4}{0.4 - x}$
$0.01 = \frac{2.303}{230} \log \frac{0.4}{0.4 - x} = 0.01 \log \frac{0.4}{0.4 - x}$
- $\frac{0.01}{0.01} = \log \frac{0.4}{0.4 - x} \Rightarrow 1 = \log \frac{0.4}{0.4 - x}$
Antilog of $1 = \frac{0.4}{0.4 - x}$
$\Rightarrow 10 = \frac{0.4}{0.4 - x} \Rightarrow x = 0.36$
For the given reaction,
$\begin{array}{lcccc} A (g) \overset{Δ}{\to} & P (g) + Q (g) + & R (g) \\ Initial & 0.4 & 0 & 0 & 0 \\ Final & 0.4 - 0.36 & 0.36 & 0.36 & 0.36 \end{array}$
Total pressure $= (0.4 - 0.36) + (3 \times 0.36)$ $= 1.12 atm$

Karnataka CET 2014

CHEMISTRY(KCET)

285442 $100 {cm}^{3}$ of $1 M {CH}_{3} COOH$ was mixed with 100 ${cm}^{3}$ of $2 M {CH}_{3} OH$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

1 4 times

2 0.25 times

3 2 times

4 0.5 times.

Explanation:

(b) ${CH}_{3} COOH + {CH}_{3} OH \to$
${CH}_{3} {COOCH}_{3} + H_{2} O$
$r_{1} = k [{CH}_{3} COOH] [{CH}_{3} OH] = k (1) (2) = 2 k$
When solution is diluted with equal volume of water, concentration is reduced to half.
$∴ r_{2} = k (\frac{1}{2}) (\frac{2}{2}) = \frac{k}{2}$
$\frac{r_{1}}{r_{2}} = \frac{2 k}{k / 2} = 4 \Rightarrow r_{2} = \frac{r_{1}}{4} = 0.25 r_{1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285443 In a first order reaction, the concentration of the reactant is reduced to $12.5 %$ in one hour. When was it half completed?

1 20 min

2 15 min

3 3 hr

4 30 min

Explanation:

(a) For a first order reaction,
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{60} \log \frac{100}{12.5}$
$= 0.0346 \min^{- 1}$
Now $t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0346} = 20 \min$

Karnataka CET 2015

CHEMISTRY(KCET)

285444 Half-life period of a first order reaction is 10 min . Starting with initial concentration 12 M , the rate after 20 min is

1

0.693 \times 3 M \min^{- 1}

2

0.0693 \times 4 \min^{- 1}

3

0.0693 M \min^{- 1}

4

0.0693 \times 3 M \min^{- 1}

Explanation:

(d) $12 M \overset{t_{1 / 2}}{\to} 6 M \overset{t_{1 / 2}}{\to} 3 M$
Initial conc.
$T_{1 / 2} = 10 \min$
$k = \frac{0.693}{10} = 0.0693 \min^{- 1}$
As $t_{1 / 2}$ is 10 min , after 20 minutes the concentration will be 3 M .
Hence, Rate $= 0.0693 \times 3 M \min^{- 1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285445 $A (g) \overset{Δ}{\to} P (g) + Q (g) + R (g)$ , follows first order kinetics with a half-life of 69.3 s at $500^{\circ} C$ . Starting from the gas 'A' enclosed in a container at $500^{\circ} C$ and at a pressure of 0.4 atm , the total pressure of the system after 230 s will be

1 1.15 atm

2 1.32 atm

3 1.22 atm

4 1.12 atm

Explanation:

(d) $T_{1 / 2} = 69.3 s, t = 230 s, a = 0.4 atm$ ,
For first order reaction,
$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{69.3} = 0.01 s^{- 1}$
$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{230} \log \frac{0.4}{0.4 - x}$
$0.01 = \frac{2.303}{230} \log \frac{0.4}{0.4 - x} = 0.01 \log \frac{0.4}{0.4 - x}$
- $\frac{0.01}{0.01} = \log \frac{0.4}{0.4 - x} \Rightarrow 1 = \log \frac{0.4}{0.4 - x}$
Antilog of $1 = \frac{0.4}{0.4 - x}$
$\Rightarrow 10 = \frac{0.4}{0.4 - x} \Rightarrow x = 0.36$
For the given reaction,
$\begin{array}{lcccc} A (g) \overset{Δ}{\to} & P (g) + Q (g) + & R (g) \\ Initial & 0.4 & 0 & 0 & 0 \\ Final & 0.4 - 0.36 & 0.36 & 0.36 & 0.36 \end{array}$
Total pressure $= (0.4 - 0.36) + (3 \times 0.36)$ $= 1.12 atm$

Karnataka CET 2014

CHEMISTRY(KCET)

285442 $100 {cm}^{3}$ of $1 M {CH}_{3} COOH$ was mixed with 100 ${cm}^{3}$ of $2 M {CH}_{3} OH$ to form an ester. The change in the initial rate if each solution is diluted with equal volume of water would be

1 4 times

2 0.25 times

3 2 times

4 0.5 times.

Explanation:

(b) ${CH}_{3} COOH + {CH}_{3} OH \to$
${CH}_{3} {COOCH}_{3} + H_{2} O$
$r_{1} = k [{CH}_{3} COOH] [{CH}_{3} OH] = k (1) (2) = 2 k$
When solution is diluted with equal volume of water, concentration is reduced to half.
$∴ r_{2} = k (\frac{1}{2}) (\frac{2}{2}) = \frac{k}{2}$
$\frac{r_{1}}{r_{2}} = \frac{2 k}{k / 2} = 4 \Rightarrow r_{2} = \frac{r_{1}}{4} = 0.25 r_{1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285443 In a first order reaction, the concentration of the reactant is reduced to $12.5 %$ in one hour. When was it half completed?

1 20 min

2 15 min

3 3 hr

4 30 min

Explanation:

(a) For a first order reaction,
$k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{60} \log \frac{100}{12.5}$
$= 0.0346 \min^{- 1}$
Now $t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{0.0346} = 20 \min$

Karnataka CET 2015

CHEMISTRY(KCET)

285444 Half-life period of a first order reaction is 10 min . Starting with initial concentration 12 M , the rate after 20 min is

1

0.693 \times 3 M \min^{- 1}

2

0.0693 \times 4 \min^{- 1}

3

0.0693 M \min^{- 1}

4

0.0693 \times 3 M \min^{- 1}

Explanation:

(d) $12 M \overset{t_{1 / 2}}{\to} 6 M \overset{t_{1 / 2}}{\to} 3 M$
Initial conc.
$T_{1 / 2} = 10 \min$
$k = \frac{0.693}{10} = 0.0693 \min^{- 1}$
As $t_{1 / 2}$ is 10 min , after 20 minutes the concentration will be 3 M .
Hence, Rate $= 0.0693 \times 3 M \min^{- 1}$

Karnataka CET 2015

CHEMISTRY(KCET)

285445 $A (g) \overset{Δ}{\to} P (g) + Q (g) + R (g)$ , follows first order kinetics with a half-life of 69.3 s at $500^{\circ} C$ . Starting from the gas 'A' enclosed in a container at $500^{\circ} C$ and at a pressure of 0.4 atm , the total pressure of the system after 230 s will be

1 1.15 atm

2 1.32 atm

3 1.22 atm

4 1.12 atm

Explanation:

(d) $T_{1 / 2} = 69.3 s, t = 230 s, a = 0.4 atm$ ,
For first order reaction,
$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{69.3} = 0.01 s^{- 1}$
$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{230} \log \frac{0.4}{0.4 - x}$
$0.01 = \frac{2.303}{230} \log \frac{0.4}{0.4 - x} = 0.01 \log \frac{0.4}{0.4 - x}$
- $\frac{0.01}{0.01} = \log \frac{0.4}{0.4 - x} \Rightarrow 1 = \log \frac{0.4}{0.4 - x}$
Antilog of $1 = \frac{0.4}{0.4 - x}$
$\Rightarrow 10 = \frac{0.4}{0.4 - x} \Rightarrow x = 0.36$
For the given reaction,
$\begin{array}{lcccc} A (g) \overset{Δ}{\to} & P (g) + Q (g) + & R (g) \\ Initial & 0.4 & 0 & 0 & 0 \\ Final & 0.4 - 0.36 & 0.36 & 0.36 & 0.36 \end{array}$
Total pressure $= (0.4 - 0.36) + (3 \times 0.36)$ $= 1.12 atm$

Karnataka CET 2014