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 $



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