Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
- Endothermic reaction means the reactant side takes heat from surrounding and get decomposed i.e ∆H=-ve
- If the equation is exothermic then it means the reactant is happy to decompose .But it's not as it's endothermic
Now
- HgO is Omitted from our solution option.
Hg is a atom so no bonds hence no bond strength occurs.
- O_2 is a molecule and so it's our answer .
thx but it's actually 143
6 CO₂ + 6 H₂O (chlorophyll + sunlight) ⇒ C₆H₁₂O₆ + 6 O₂
the reaction takes place in presence of sunlight and chlorophyll ..
CO₂ is oxidized and forms Glucose . And oxygen is evolved in this process.
