Question
Given
C(grahite) + O2(g) → CO2(g)
ΔrH° = - 393.5 kJ mol-1
H2(g) + 1/2O2(g) → H2O (l)
ΔrH° = +890.3 kJ mol-1
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction
C(grahite) + 2H2(g) →CH4 will be
-
+74.8 kJ mol–1
-
+144.0 kJ mol–1
-
–74.8 kJ mol–1
-
–144.0 kJ mol–1
Solution
C.
–74.8 kJ mol–1
CO2(g) + 2H2O (l)→ CH4 (g) + 2O2(g);
ΔrH°= 890.3
ΔfH° –393.5 –285.8 ? 0
ΔrH°= Σ(ΔfH°)products -Σ(ΔfH°)reactant
890.3 = [ 1 x(ΔfH°)CH4 + 2x0]-[1x(-393.5)+2(-285.8)]
(ΔfH°)CH4 = 890.3-965.1 = -74.8 kJ/mol
