Lighters are usually fueled by butane (C4H10). When 1 mole of butane burns at constant pressure it produces 2658kJ of heat and does 3kJ of w?

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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

  • 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.

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