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What is the amount of work done when 0.5 mole of methane, `CH_(4)`(g) is subjected to combustion at 300 K? (given, `R=8.314 Jk^(-1) mol^(-1))`A. `-2494 J`B. `-4988 J`C. `+4988 J`D. `+2494 J` |
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Answer» Correct Answer - D `CH_(4)(g) + 2CO_(2)(g) to CO_(2)(g) + 2H_(2)O(l)` `W =Deltan RT` 1 mole of reacts with 2 moles of oxygen to produce 2 moles of `CO_(2)` Hence, `0.5` mole of `CH_(4)` reacts with 1 mole of `O_(2)` to produce 1 mole of `CO_(2)` `therefore n_(1)`= moles of reactants =1.5 `n_(2)` =moles of product =1 `Deltan=n_(2) -n_(1) = 1 - 1.5 =-0.5` `W= -(-0.5) xx 300 xx 8.314 J K^(-1) mol^(-1)` `=+2494` J |
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