Question

Tolong yang bisa no 24&25

Answer 1

24. mol  CH4 = 6 ÷ 16 = 0,375 dengan ΔH = - 225 kJ
maka ΔH pembakaran standar CH4 (dalam 1 mol) = 1/0,375 × - 225 = - 600 kJ
jawaban: A

25.
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O  ΔH = - 2826,6 kJ

ΔH = [(6 × ΔHf CO2) + (6 × ΔHf H2O)] - [(ΔHf C6H12O6 - (6 × ΔHf O2)]
- 2826,6 = [(6 × - 393,3) + (6 × - 241,8)] - [ΔHf C6H12O6 - (6 × 0)]
ΔHf C6H12O6 = - 984 kJ
jawaban: A