# Hess Law Homework help?

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1: 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=?

2: Consider the reaction

N2(g)+3H2(g)→2NH3(g), ΔH =−92.3kJ

What will ΔH be for the reaction if it is reversed?

3: calculate the reaction enthalpy, ΔH, for the following reaction:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)

Use the series of reactions that follow:

C(s)+2H2(g)→CH4(g), ΔH =−74.8 kJ.

C(s)+O2(g)→CO2(g), ΔH =−393.5 kJ.

2H2(g)+O2(g)→2H2O(g), ΔH =−484.0 kJ.

H2O(l)→H2O(g), ΔH =44.0 kJ.

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• Well with the first question everything was multiplied by four so the ΔH will be four times as much. Think of it like you ran the experiment four times, each time the ΔH was -2043kJ so 4 × -2043kJ = -8172kJ.

If you reverse a reaction you multiply the ΔH by -1. You can think of like you have to the exact opposite of what the forward reaction is. If the forward reaction requires 92.3kJ then the reverse releases 92.3kJ, so ΔH = 92.3kJ.

Number three is a tricky problem. You have to try and get the reaction you want from simpler reactions.

Reverse the first reaction

CH4(g) → C(s) + 2H2(g) , ΔH = 74.8kJ

Add the second reaction to the line above

CH4(g) + O2(g) → 2H2(g) + CO2(g) , ΔH = -318.7kJ

Note that the C(s) canceled out. Add the third reaction to the line above

CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) , ΔH = -802.7kJ

Note that the 2H2(g) canceled out. Reverse the fourth reaction and add two of it.

CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) , ΔH = -890.7kJ

Note that the 2H2O(g) canceled out. 