1.

What is Deltan_(g) for the combusion of 1 mole of benzene, when both reactants and products are gases at 298K

Answer»

0
1
0.5
1.5

Solution :`C_(6)H_(6)(G)+(15)/(2)O_(2)(g) to 6CO_(2)(g)+3H_(2)O(g)`
`Deltan=6+3-1-(15)/(2)=+0.5`


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