This is an incomplete question, here is a complete question.
A mole of X reacts at a constant pressure of 43.0 atm via the reaction
[tex]X(g)+4Y(g)\rightarrow 2Z(g)[/tex], 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 with the appropriate units.
Answer : The value of [tex]\Delta E[/tex] is, -61.9 kJ
Explanation:
Formula used :
[tex]\Delta H=\Delta E+P(V_2-V_1)[/tex]
[tex]\Delta H[/tex] = change in enthalpy = -75.0 kJ
[tex]\Delta E[/tex] = change in internal energy
[tex]V_1[/tex] = initial volume of gas = 5.00 L
[tex]V_2[/tex] = final volume of gas = 2.00 L
P = pressure of gas = 43.0 atm
Now put all the given values in the above formula, we get:
[tex]-75.0kJ=\Delta E+43.0atm\times (2.00-5.00)L[/tex]
[tex]-75.0kJ=\Delta E-129L.atm[/tex]
Conversion used : (1 L.atm = 101.3 J)
[tex]-75.0kJ=\Delta E-129\times 101.3J[/tex]
[tex]-75.0kJ=\Delta E-13067.7J[/tex]
[tex]-75.0kJ=\Delta E-13.07kJ[/tex]
[tex]\Delta E=-61.9kJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] is, -61.9 kJ
