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Click the DeltaH is an Extensive Property button within the activity, 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: 4C3H8(g)+20O2(g)→12CO2(g)+16H2O(g), ΔH2=?

Respuesta :

Edufirst
Calculate the change in enthalpy for Reaction 2.
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Reaction 2: 4C3H8(g)+20O2(g)→12CO2(g)+16H2O(g), ΔH2=?

An extensive property is one whose value depends on the amount of material.


The change of enthalpy is proportional to the quantity, then given that the equation 2 is the equation 1 times 4, the change of enthalpy of equation 2 is 4 times the change of enthalpy of equation 1.

ΔH2 = 4*ΔH1 = 4* (-2043 kJ) = -8172 kJ

Answer: -8172 kJ
superman1987

The answer is = -8172 kJ

The Explanation:

when Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure. It is given the symbol ΔH, read as "delta H".

So to Calculate the change in enthalpy for Reaction 2, we need to understand

first:

that the first reaction is the same as he second one but we assume that we multiply the reaction 2 by 4 to be like that:

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

Reaction (2): 4C3H8(g)+20O2(g)→12CO2(g)+16H2O(g) So ΔH2=?

when The change of enthalpy is proportional to the quantity. and here he reaction 2 is = reaction 1 * 4.

So the enthalpy change for the reaction 2 will be:

ΔH2 = 4*ΔH1 = 4* (-2043 kJ) = -8172 kJ

∴ The answer is -8172 KJ


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