1
MHT CET 2023 11th May Evening Shift
+1
-0

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

A
0.11 minute$$^{-1}$$
B
0.22 minute$$^{-1}$$
C
0.34 minute$$^{-1}$$
D
0.42 minute$$^{-1}$$
2
MHT CET 2023 11th May Evening Shift
+1
-0

Identify the expression for average rate for following reaction.

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

A
$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=-\frac{1}{3} \frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
B
$$-\frac{1}{3} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
C
$$\frac{-\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
D
$$-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2\right]}{\Delta \mathrm{t}}=\frac{-\Delta\left[\mathrm{H}_2\right]}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{NH}_3\right]}{\Delta \mathrm{t}}$$
3
MHT CET 2023 11th May Morning Shift
+1
-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:

A
$$1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$
B
$$2.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$
C
$$5.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$
D
$$8.66 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$
4
MHT CET 2023 11th May Morning Shift
+1
-0

Slope of the graph between $$\log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$$ (y axis) and time ( $$x$$ axis) for first order reaction is equal to:

A
$$+\frac{\mathrm{k}}{2.303}$$
B
$$\mathrm{k}$$
C
$$-\mathrm{k}$$
D
$$-\frac{2.303}{\mathrm{k}}$$
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Medical
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