Answer:
7. A) I, II
; 8. D) 2.34e9 kJ
Step-by-step explanation:
7. Combustion of ethanol
I. The negative sign for ΔH shows that the reaction is exothermic.
II. The enthalpy change would be different if gaseous water were produced.
That's because it takes energy to convert liquid water to gaseous water, and this energy is included in the value of ΔH.
III. The reaction is a redox reaction, because
- Oxygen is reacting with a compound
- The oxidation number of C increases
- The oxidation number of O decreases.
IV. The products of the reaction occupy a smaller volume than the reactants, because 3 mol of gaseous reactant are forming 2 mol of gaseous product.
Therefore, only I and II are correct.
7. Hindenburg
Data:
V = 2.00 × 10⁸ L
p = 1.00 atm
T = 25.1 °C
ΔH = -286 kJ·mol⁻¹
Calculations:
(a) Convert temperature to kelvins
T = (25.1 + 273.15) K = 298.25 K
(b) Moles of hydrogen
Use the <em>Ideal Gas Law</em>:
pV = nRT
n = (pV)/(RT)
n = (1.00 × 2.00 × 10⁸)/(0.082 06 × 298.25) = 8.172 × 10⁶ mol
(c) Heat evolved
q = nΔH = 8.172 × 10⁶ × (-286) = -2.34 × 10⁹ kJ
The hydrogen in the Hindenburg released 2.34e9 kJ
.
