Question

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. 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 -194. kJ of work is done on the mixture during the reaction. a.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 3 significant digits b. Is the reaction exothermic or endothermic?

Answer 1

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative

Step-by-step explanation:

given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ

then,

change in enthalpy (ΔH) = -162( released energy)

work(w) = -194KJ

change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.

At constant pressure the change in enthalpy is given as

ΔH = ΔU + PΔV

ΔU = change in energy

ΔV = change in volume

P = pressure

w = -pΔV

therefore,

ΔH = ΔU -W

to evaluate energy change we have,

ΔU =ΔH + W

ΔU = -162+ (-194KJ)

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative