Question

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 215. 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 -155. kJ of work is done on the mixture during the reaction.

Calculate the change in energy of the gas mixture during the reaction. Round your answer to 2 significant digits.

Is the reaction exothermic or endothermic?

Answer 1

a) ΔU = 370 KJ

b) Endothermic

Step-by-step explanation:

a)

The change in energy of the mixture can be given by first law of thermodynamics as:

ΔQ = ΔU + W

ΔU = ΔQ - W

where,

ΔQ = change in heat energy of system

ΔU = Change in internal energy of gases

W = Work done on gases = - 155 KJ

For an isobaric process (i.e constant pressure) we know that:

ΔQ = change in enthalpy = ΔH

ΔQ = 215 KJ

Therefore, using values in the equation, we get:

ΔU = 215 KJ - (-155 KJ)

ΔU = 370 KJ

b)

Since, the enthalpy of products is greater than the reactants. Therefore, this is an endothermic reaction.