MHT CET 2023 11th May Evening Shift | Chemical Kinetics Question 110 | Chemistry

For a reaction $$\mathrm{A}+\mathrm{B} \rightarrow$$ product, if $$[\mathrm{A}]$$ is doubled keeping $$[\mathrm{B}]$$ constant, the rate of reaction doubles. Calculate the order of reaction with respect to A.

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MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

Calculate the rate constant of the first order reaction if $$80 \%$$ of the reactant reacted in 15 minute.

0.11 minute$$^{-1}$$

0.22 minute$$^{-1}$$

0.34 minute$$^{-1}$$

0.42 minute$$^{-1}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

Identify the expression for average rate for following reaction.

$$\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$$

$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=-\frac{1}{3} \frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$

$$-\frac{1}{3} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$

$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$

$$-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$

MHT CET 2023 11th May Morning Shift

MCQ (Single Correct Answer)

-0

For an elementary reaction

$$2 \mathrm{~A}+\mathrm{B} \longrightarrow 3 \mathrm{C}$$

rate of appearance of $$\mathrm{C}$$ is $$1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$, the rate of disappearance of $$\mathrm{A}$$ is:

$$1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$

$$2.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$

$$5.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$

$$8.66 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$

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