- A$\mathrm{CoCl}_{3}$
- B$\mathrm{FeCl}_{2}$
- C$\mathrm{CoCl}_{2}$
- ✓$\mathrm{FeCl}_{3}$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$Prussian blue
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$N{H_3} + {O_2}\xrightarrow[{Pt}]{\Delta }x + {H_2}O$
${x} + {O_2} \to y$
$y + {H_2}O{\text{(excess)}} \to {\text{x + z}}$
The incorrect option is

$\mathrm{A}+\mathrm{B} \underset{\text { Step } 3}{\text { Step } 1} \mathrm{C} \xrightarrow{\text { Step } 2} \mathrm{P}$
Some details of the above reaction are listed below.
| Step |
Rate constant $\left(\sec ^{-1}\right)$ |
Activation energy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ |
| $1$ | ${k}_1$ | $300$ |
| $2$ | ${k}_2$ | $200$ |
| $3$ | ${k}_3$ | $\mathrm{Ea}_3$ |
If the overall rate constant of the above transformation (k) is given as $\mathrm{k}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_3}$ and the overall activation energy $\left(E_2\right)$ is $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then the value of $\mathrm{Ea}_3$ is $\qquad$ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer)