Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
Given that
Mass of water = 65.34 g
Amount of heat = mass of water * specific heat (temperature change
)
= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C
= 907.63 J
= 0.908 KJ
And
1 cal = 4.186798 J
907.63 J * 1 cal / 4.186798 J =216.78 cal
Or0.218 kcal
1 is D - double-replacements do not make solid metals
2 is A - to have complete combustion the original compound must ONLY have C, H and O
3 is B - the elemental Mg replaces the H in the HCl
Answer:
Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide
Explanation:
