When 20.00 moles of H₂(g)reacts with 10.00 mol of O₂(g)to form 20.00 mol of H₂O(l)at 25°C and a constant pressure of 1.00 atm.If 1366 kJ of heat are released during this reaction,and PΔV is equal to -74.00 kJ,then A)ΔH° = +1366 kJ and ΔE° = +1440 kJ. B)ΔH° = +1366 kJ and ΔE° = +1292 kJ. C)ΔH° = -1366 kJ and ΔE° = -1292 kJ. D)ΔH° = -1366 kJ and ΔE° = -1440 kJ.

Question

Quizplus · Accepted Answer

This is a thermochemical equation where we need to find the standard enthalpy change and the standard internal energy change. We can start by using the equation:

ΔH° = ΔE° + PΔV

where ΔH° is the standard enthalpy change, ΔE° is the standard internal energy change, and PΔV is the work done by the system.

We are given that 1366 kJ of heat are released during the reaction and PΔV is -74.00 kJ. Substituting these values in the above equation gives:

ΔH° = -1366 kJ

To find ΔE°, we can use the equation:

ΔE° = ΔH° - PΔV

Substituting the values we get:

ΔE° = -1366 kJ - (-74.00 kJ) = -1292 kJ

Therefore, the correct answer is C as it matches both the enthalpy and internal energy values.

When 2000 Moles of H2(g)reacts with 10

When 2000 Moles of H₂(g)reacts with 10