Answer:
ΔH°f P4O10(s) = - 3115.795 KJ/mol
Explanation:
- P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
- ΔH°rxn = ∑νiΔH°fi
∴ ΔH°rxn = - 327.2 KJ
∴ ΔH°f H2O(l) = - 285.84 KJ/mol
∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol
⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ
⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2
⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol
Answer:
Q = 28.9 kJ
Explanation:
Given that,
Mass of Aluminium, m = 460 g
Initial temperature, 
Final temperature, 
We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :
So, 28.9 kJ of heat is required to raise the temperature.
Answer : The value of 
for this reaction is 36.18 kJ
Explanation :
First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.
As per first law of thermodynamic,
where,
= internal energy of the system
q = heat added or rejected by the system
w = work done
As we are given that:
q = 38.65 kJ
w = -2.47 kJ (system work done on surrounding)
Now put all the given values in the above expression, we get:
Therefore, the value of for this reaction is 36.18 kJ
Answer:
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Hope this helps
Excerpt from textbook
