MHT CET 2024 9th May Evening Shift | Chemical Kinetics Question 23 | Chemistry

For the reaction, $\mathrm{N}_{2(\mathrm{g})}+3 \mathrm{H}_{2(\mathrm{g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{g})}$ $\mathrm{NH}_3$ is formed at a rate of $0.088 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$. Calculate consumption rate of $\mathrm{N}_{2(\mathrm{g})}$.

$0.011 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$0.022 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$0.033 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$0.044 \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

Consider the reaction $3 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \longrightarrow \mathrm{I}_3^{-}+2 \mathrm{SO}_4^{2-}$, at a particular time t , $\frac{\mathrm{d}\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{dt}}$ is $2.2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. What is the value of $\frac{\mathrm{d}\left[\mathrm{S}_2 \mathrm{O}_8^{2-}\right]}{\mathrm{dt}}$ ?

$1.1 \times 10^{-2} \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$2.2 \times 10^{-2} \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$3.3 \times 10^{-2} \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

$6.6 \times 10^{-2} \mathrm{~mol} \mathrm{~dm}^{-3} \mathrm{~s}^{-1}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

For the reaction, $\mathrm{NO}_{2(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})} \longrightarrow \mathrm{NO}_{(\mathrm{g})}+\mathrm{CO}_{2(\mathrm{~g})}$ rate of reaction is proportional to square of $\left[\mathrm{NO}_2\right]$ and independent of [CO]. What is the rate law equation?

$\mathrm{r}=\mathrm{k} \frac{\left[\mathrm{NO}_2\right]^{\frac{1}{2}}}{[\mathrm{CO}]}$

$\mathrm{r}=\mathrm{k}\left[\mathrm{NO}_2\right]^2[\mathrm{CO}]^0$

$\mathrm{r}=\frac{1}{2} \mathrm{k}\left[\mathrm{NO}_2\right][\mathrm{CO}]$

$\mathrm{r}=\mathrm{k}\left[\mathrm{NO}_2\right]^2[\mathrm{CO}]$

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