Question

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.

Answer 1

ΔE = -2661 KJ/mole

ΔH = -2658 KJ/mole

Step-by-step explanation:

ΔH = q - PΔV

ΔE = q + w

First, to find ΔE:

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 ΔE is negative.

ΔE = -2661 KJ/mole

Second, to find ΔH:

ΔH = q - PΔV

ΔH = 2658 kJ(q) - PΔV

Now, the question states that butane burns at a constant pressure; that just translates to the pressure of the reaction is equal to 0.

ΔH = 2658 KJ(q) - (0)ΔV

ΔH = 2658 KJ - 0

ΔH = 2658 kJ, BUT, like before, the reaction PRODUCES heat, which also mean ΔH is negative.

ΔH = -2658 KJ/mole

I hope this helped! Have a nice week.