Chemical Thermodynamics Question 491
Question: The $ \Delta H_{F}^{O} $ for $ CO_2(g),CO(g) $ and $ H_2O(g) $ are $ -,393.5,,-110.5 $ and $ -,241.8kJmol $ , respectively. The standard enthalpy change (in kJ) for the reaction $ C{O_{2(g)}}+H_2{(g)}\xrightarrow{{}} C{O_{(g)}}+H_2{O _{(g)}} $ is:
Options:
A) + 524.1
B) + 41.2
C) - 262.5
D) - 41.2
Show Answer
Answer:
Correct Answer: B
Solution:
- (i) $ {C_{( s )}}+{O_{2(g)}}\to C{O_{2(g)}};\Delta H=-393.5kJ/mol $
(ii) $ {C_{( s )}}+\frac{1}{2}{O_{2(g)}}\to C{O_{(g)}};\Delta H=-110.5kJ/mol $
(iii) $ {H_{2( g )}}+\frac{1}{2}{O_{2(g)}}\to H_2{O_{(g)}};\Delta H=-241.8kJ/mol $
For getting $ C{O_{2( g )}}+{H_{2(g)}}\to C{O_{(g)}}+H_2{O_{(g)}}, $ add (ii) and (iii) and subtract (i). Thus, - 110.5 - 241.8 + 393.5 = 395.3 = 41.2 kJ/mol
