Click the The button labeled delta H is an Extensive Property. button, and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the change in enthalpy for a reaction, ΔH, is an extensive property. Using this property, calculate the change in enthalpy for Reaction 2. Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Reaction 2: 5C3H8(g)+25O2(g)→15CO2(g)+20H2O(g), ΔH2=?

ΔH = - 10215 KJ

Explanation:

An extensive property is the property of a substance that changes with the mass of the substance. So, if we increase the mass by some factor than the extensive properties of that system shall increase by that factor as well. We are given two reaction as follows:

Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ

Reaction 2: 5C3H8(g)+25O2(g)→15CO2(g)+20H2O(g), ΔH2=?

If we take 5 common from reaction 2 then it becomes:

5[C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)], ΔH2=?

5(Reaction 1)

This shows that reaction 2 is equivalent to the reaction 1, it just has 5 times the mass of reaction 1. Since, change in enthalpy is an extensive property. Therefore, it will also become 5 times of reaction 1:

ΔH = 5(-2043 KJ)

ΔH = - 10215 KJ

58 km per hour : d

c. 75.7

i think that is the right