Question

A mole of X reacts at a constant pressure of 43.0 atm via the reaction

X(g)+4Y(g)→2Z(g),      ΔH∘=−75.0 kJ

Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.

Express your answer numerically in kilojoules.

Answer 1

The following quantities are given

ΔH0 = -75.0 kJ , Pressure, P = 43.0 atm = 43 atm * 101325 Pa/1 atm = 4356975 Pa = 4356975 N / m2

Volume change, ΔV = Final - Initial = 2.0 L - 5.0 L = -3.0 L = -3.0 *10-3 m3

Work done, w = - PΔV
= - (4356975 N / m2) (-3.0 *10-3 m3)
= 13070 N m
= 13070 J
= 13.07 kJ
We know that at consatnt pressure q_p = ΔH0 = -75.0 kJ
Also ΔE = q + w
= - 75.0 kJ + 13.07 kJ
= -61.93 kJ