Chemical Thermodynamics Question 421
Question: Standard enthalpy of combustion of $ CH_3 $ is $ -,890kJmo{{l}^{-1}} $ and standard enthalpy of vaporisation of water is $ 40.5kJmo{{l}^{-1}} $ . The enthalpy change of the reaction $ CH_4( g )+2O_2( g )\xrightarrow{{}}CO_2( g )+H_2O( g ) $
Options:
A) $ -809.5kJmo{{l}^{-1}} $
B) $ -890kJmo{{l}^{-1}} $
C) $ 809kJ,mo{{l}^{-1}} $
D) $ -971kJ,mo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ CH_4( g )+2O_2( g )\xrightarrow{{}}CO_2( g )+2H_2O( l ) $
$ \Delta H=-890kJ $ …. (i) $ 2H_2O( l )\xrightarrow{{}}2H_2O( g ); $
$ \Delta H=2\times 40.5=81kJ $ …. (ii) From (i) + (ii), we get $ CH_4(g)+2O_2(g)\xrightarrow{{}}CO_2( g )+2H_2O( g ) $
$ \Delta H=-890+81=-809kJ $
