COMEDK 2024 Afternoon Shift | Chemical Kinetics Question 6 | Chemistry

The temperature $$(\mathrm{T})$$ and rate constant $$(\mathrm{k})$$ for a first order reaction $$\mathrm{R} \rightarrow \mathrm{P}$$, was found to follow the equation $$\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+8.0$$. The pre-exponential factor '$$\mathrm{A}$$' and activation energy $$\mathrm{E}_{\mathrm{a}}$$, respectively are: [Given: $$\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$]

$$ 8.0 \mathrm{~s}^{-1} \text { and } 16.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

$$ 6.0 \mathrm{~s}^{-1} \text { and } 10^8 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

$$ 1 \times 10^8 \mathrm{~s}^{-1} \text { and } 38.3 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

$$ 1 \times 10^{-8} \mathrm{~s}^{-1} \text { and } 16.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

Given below a first order reaction in the gas phase

$$\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$$

If the initial pressure of the system is $$\mathrm{P}_{\mathrm{i}}$$ and the total pressure at $$\mathrm{t}$$ seconds is $$\mathrm{P}_{\mathrm{t}}$$, the rate constant $$\mathrm{k}$$ for the reaction is:

$$ \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{t}}} $$

$$ \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)} $$

$$ \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)} $$

$$ \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(\mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)} $$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

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For the reaction, $$\mathrm{A}+3 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$$, the concentration of $$\mathrm{A}$$ changes from 0.0150 to 0.0125 in 1 minute. The rate of formation of $$\mathrm{C}$$ in $$\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$ is:

$$6.32 \times 10^{-5}$$

$$8.32 \times 10^{-5}$$

$$3.26 \times 10^{-5}$$

$$2.5 \times 10^{-5}$$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

The rate of appearance of bromine is related to the disappearance of bromide ion in the equation given below is:

$$\mathrm{BrO}_3^{-} \text {(aq) }+5 \mathrm{Br}^{-} \text {(aq) }+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_2(\mathrm{l})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$$

$$ \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=\frac{5}{3} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}} $$

$$ \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=-\frac{1}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}} $$

$$ \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}} $$

$$ \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}} $$

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