At a constant pressure of 1.50 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. )

Question

Quizplus · Accepted Answer

The enthalpy change can be calculated using the equation:
ΔH = ΔE + PΔV
where P is the constant pressure and ΔV is the volume change.

Substituting the given values, we get:
ΔH = 58.0 kJ + (1.50 atm)(12.6 L)(101.325 J/L·atm)/(1000 J/kJ)
ΔH = 59.9 kJ

Therefore, the enthalpy change is 59.9 kJ, which corresponds to choice C.

At a Constant Pressure of 1