Chemical Thermodynamics Question 161
Question: The value of enthalpy change ( $ \Delta H $ ) for the reaction $ C_2H_2OH(l)+3O_2(g)\to 2CO_2(g)+3H_2O(l) $ a $ 27{}^\circ C $ is -1366.5 kJ $ mo{{l}^{-1}} $ . The value of internal energy change for the above reaction at this temperature will be
Options:
A) -1371.5 kJ
B) -1369.0 kJ
C) -1364.0 kJ
D) -1361.5 kJ
Show Answer
Answer:
Correct Answer: C
Solution:
- Relation between $ \Delta H $ (enthalpy change) and $ \Delta E $ (internal energy change) is $ \Delta H=\Delta E+\Delta n_{g}RT $ Where $ \Delta n_{g} $ =(moles of gaseous products)-(moles of gaseous reactants) For the given reaction, $ \Delta n_{g}=2-3=-1 $
$ \Rightarrow -1366.5=\Delta E-1\times 8.314\times {{10}^{-3}}\times 300 $
$ \therefore \Delta E=-1364.0kJ,mo{{l}^{-1}} $
BETA
