Lighters are usually fueled by butane (C4H10). When 1 mole of butane burns at constant pressure, it produces 2658 kJ of heat and does 3 kJ of work.
What are the values of ΔH and ΔE for the combustion of one mole of butane?
Could you please explain this
1 Answer
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So, consider the necessary formulas:
deltaH = q – PdeltaV
deltaE = q + w
First, to find delta E:
The reaction PRODUCES 2658 kJ of h (q), and does 3 kJ of work (w).
2658 kJ(q) + 3 kJ(w) = 2661 kJ, BUT the reaction PRODUCES heat, which means deltaE is negative.
deltaE = -2661
Second, to find deltaH:
deltaH = q – PdeltaV
deltaH = 2658 kJ(q) – PdeltaV
Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.
deltaH = 2658 kJ(q) – (0)deltaV
deltaH = 2658 kJ(q) – 0
deltaH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean deltaH is negative.
deltaH = -2658 kJ
I hope this helped! Have a nice week.
