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What is `Deltan_(g)` for the combusion of 1 mole of benzene, when both reactants and products are gases at 298K |
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Answer» Correct Answer - C `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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