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

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}.