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.
Please help me with these problems. I am beyond confused to this concept.
1 Answer
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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.
