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
In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
Read the paper more carefully if you don't understand it
Answer:
= -457.9 kJ and reaction is product favored.
Explanation:
The given reaction is associated with 2 moles of 
Standard free energy change of the reaction () is given as:
, where T represents temperature in kelvin scale
So, 
So, for the reaction of 1.57 moles of , 
As, is negative therefore reaction is product favored under standard condition.
Answer:
-250.3kJ
Explanation:
Based in the reactions and using -<em>Hess's law-</em>:
(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ
(2) 4PCl₅(g) → P₄(s) + 10Cl₂ ΔH₂ = 3438kJ
The sum of (1) + (2) is:
4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ
Dividing this reaction in 4:
PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>
