Question

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 243.kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that −174.kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.

Answer 1

Answer: The change of energy of the gas mixture during the reaction in 69 kJ

Explanation:

According to first law of thermodynamics:

$\Delta E=q+w$

$\Delta E$=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system

w = -174 kJ

q = +243 kJ   {Heat has increased so heat is absorbed}

$\Delta E=-174kJ+(+243)kJ=69kJ$

Thus the change of energy of the gas mixture during the reaction in kJ is 69