Chemical Thermodynamics Question 375
Question: What is $ \Delta n_{gas} $ for the combustion of 1 mole of benzene, when both the reactants and the products are at 298 K -
Options:
A) 0
B) 1/2
C) 3/2
D) $ -3/2 $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ C_6H_6(l)+\frac{15}{2}O_2(g)\to 6CO_2(g)+3H_2O(l) $
$ \therefore \Delta n=6-15/2=-3/2 $
