Chemistry - Thermochemistry Question with Solution | TestHub
When 3.0 g graphite is burnt in limited supply of oxygen at 298 K and 1 bar, 7.5 kcal heat is released. No solid is left and the gaseous product formed is not absorbed in aqueous KOH solution. When 4.0 g graphite is burnt in excess supply of oxygen at 298 K and 1 bar, 32 kcal heat is released. No solid residue is left and the gaseous product formed is absorbed completely in aqueous KOH solution. When 4.0 g CO2(g) is decomposed completely into CO(g) and O2(g) at 29 K and 1 bar, the enthalpy of system increases (in kcal) by
Answer:
Solution:
The product not absorbed in KOH is CO. The reaction is:
C + 1/2O₂ tarrow CO
The enthalpy change for this reaction is Δ H = -7.5 × 12/3.0 = -30 kcal/mol.
The absorbed product is CO₂. The reaction is:
C + O₂ tarrow CO₂
The enthalpy change for this reaction is Δ H = -32 × 12/4.0 = -96 kcal/mol.
Now, consider the reaction:
CO₂ tarrow CO + 1/2O₂
The enthalpy change for this reaction is Δ H = +96 - 30 = +66 kcal/mol.
For 4.0 g of CO₂, which is 4.0/44 mol, the enthalpy change is:
Δ H = 66 × 4.0/44 = 6 kcal
The answer is 6.
We assume the enthalpies are temperature-independent, so the answer remains unchanged at 298 K.
