At a constant pressure of 2.50 atm what is the enthalpy change ΔH for a reaction where the internal energy change ΔE is 58.0 kJ and the volume increase is 12.6 L? (1 L·atm = 101.325 J. )

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Quizplus · Accepted Answer

The enthalpy change ΔH can be calculated using the formula ΔH = ΔE + PΔV, where ΔE is the change in internal energy, P is the pressure, and ΔV is the change in volume. Given ΔE = 58.0 kJ, P = 2.50 atm, and ΔV = 12.6 L, first convert the volume change to the same units as the energy change using 1 L·atm = 101.325 J. Thus, PΔV = 2.50 atm * 12.6 L * 101.325 J/L·atm = 3183.225 J = 3.183 kJ. Therefore, ΔH = 58.0 kJ + 3.183 kJ = 61.183 kJ, which rounds to 61.2 kJ.

At a Constant Pressure of 2