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Consider the reactions: (a) $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} O_{6}(a q)+6 O_{2}(g)$ (b) $O_{3}(g)+H_{2} O_{2}(l) \rightarrow H_{2} O_{(l)}+2 O_{2}(g)$ Why it is more appropriate to write these reactions as : (a) $6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} \mathrm{O}_{6}($ aq $)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}+6 \mathrm{O}_{2}(g)$ (b) $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g)$ Also suggest a technique to investigate the path of the above (a) and (b) redox reactions
Consider the reactions: (a) $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} O_{6}(a q)+6 O_{2}(g)$ (b) $O_{3}(g)+H_{2} O_{2}(l) \rightarrow H_{2} O_{(l)}+2 O_{2}(g)$ Why it is more appropriate to write these reactions as : (a) $6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} \mathrm{O}_{6}($ aq $)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}+6 \mathrm{O}_{2}(g)$ (b) $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g)$ Also suggest a technique to investigate the path of the above (a) and (b) redox reactions
CBSE | Chemistry | Class 11 | Exercise 1 | NCERT | Redox Reactions
Consider the reactions: (a) $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} O_{6}(a q)+6 O_{2}(g)$ (b) $O_{3}(g)+H_{2} O_{2}(l) \rightarrow H_{2} O_{(l)}+2 O_{2}(g)$ Why it is more appropriate to write these reactions as : (a) $6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} H_{12} \mathrm{O}_{6}($ aq $)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}+6 \mathrm{O}_{2}(g)$ (b) $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g)$ Also suggest a technique to investigate the path of the above (a) and (b) redox reactions

Solution:

(a)

Stage 1:

$\mathrm{H}_{2} \mathrm{O}$ breaks to give $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$.

$2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$

Stage 2 :

The $H_{2}$ created in before step decreases $C O_{2}$, along these lines produce glucose and water.

$6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2}(g) \rightarrow C_{6} H_{12} O_{6}(s)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}$

The net response is as given beneath:

$\left[2 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow 2 \mathrm{H}_{2}(g)+O_{2}(g)\right] \times 6$

$6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2}(g) \rightarrow C_{6} H_{12} \mathrm{O}_{6}(s)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}$

$6 \mathrm{CO}_{2}(g)+12 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow C_{6} \mathrm{H}_{12} \mathrm{O}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}_{(l)}+6 \mathrm{O}_{2}(g)$

This is the appropriate way of composing the response as the response additionally produce water particles in the photosynthesis interaction.

The way can be found with the assistance of radioactive $\mathrm{H}_{2} \mathrm{O}^{18}$ rather than $\mathrm{H}_{2} \mathrm{O}$.

(b)

Stage 1 :

$O_{2}$ is delivered from each of the reactants $O_{3}$ and $H_{2} O_{2}$. That is the explanation $O_{2}$ is composed twice.

$O_{3}$ breaks to shape $O_{2}$ and 0 .

Stage 2:

$\mathrm{H}_{2} \mathrm{O}_{2}$ responds with $\mathrm{O}$ delivered in the prior advance, along these lines produce $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{O}_{2}$.

$O_{3}(g) \rightarrow O_{2}(g)+O_{(g)}$

$\mathrm{H}_{2} \mathrm{O}_{2}(l)+\mathrm{O}_{(g)} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2}(\mathrm{~g})$

$\mathrm{H}_{2} \mathrm{O}_{2(l)}+\mathrm{O}_{3}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$

The way can be found with the assistance of $\mathrm{H}_{2} \mathrm{O}_{2}^{18}$ or $\mathrm{O}_{3}^{18}$.

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