Question

For a particular reaction, system absorbs 6 kJ of heat and does 1.5 kJ of work on its surrounding. What is enthalpy change of system?

• A. +7.5 kJ
• B. +6.0 kJ
• C. +4.5 kJ
• D. -1.5 kJ

Answer 1

Answer: (B)

+6.0 kJ

Answer 2

For a constant pressure process, the enthalpy change is given by

$$\Delta H = q_p = +6\,\text{kJ}$$

Even though the internal energy change is

$$\Delta U = q - w = 6 - 1.5 = +4.5\,\text{kJ},$$

at constant pressure, the enthalpy change is the heat absorbed, hence +6 kJ.

At constant pressure, $\Delta H = q_p$. Given $q = +6$ kJ, therefore $\Delta H = +6$ kJ.