In a three reactions system the first two reactions are endothermic with DeltaH values of 138kj and 241kj. if the final reaction is exothermic what is the minimum value of DeltaH3 that would make the overall reaction exothermic? explain your reasoning.
i beleive the minimum value of DeltaH3 would be -379 because if the final reaction is exothermic it would be releasing the heat absorbed?
I almost agree with you. However, if you add 241 and 18 and -379 you end up with zero and zero is neither endo nor exo so I would make that -380 so as to have a delta H = a negative number.
To determine the minimum value of ΔH3 that would make the overall reaction exothermic, we need to consider the signs and magnitudes of the enthalpy changes (ΔH) for each of the reactions involved.
In an exothermic reaction, the overall ΔH value will be negative, indicating that heat is released to the surroundings. On the other hand, in an endothermic reaction, the ΔH value will be positive, indicating that heat is absorbed from the surroundings.
Let's label the three reactions as follows:
Reaction 1: ΔH1 = 138 kJ (endothermic)
Reaction 2: ΔH2 = 241 kJ (endothermic)
Reaction 3: ΔH3 (unknown)
To determine the overall enthalpy change for the system, we need to sum up the individual ΔH values for all three reactions:
Overall ΔH = ΔH1 + ΔH2 + ΔH3
Since the first two reactions are endothermic and have positive ΔH values, the sum of their ΔH values will also be positive.
First, let's calculate the sum of ΔH1 and ΔH2:
ΔH1 + ΔH2 = 138 kJ + 241 kJ = 379 kJ
Since the sum of ΔH1 and ΔH2 is positive (379 kJ), the overall reaction will also be endothermic if ΔH3 is positive.
However, we want to determine the minimum value of ΔH3 that would make the overall reaction exothermic, meaning the overall ΔH value needs to be negative.
For the overall reaction to become exothermic, the value of ΔH3 should be sufficiently large and negative so that it can offset the positive sum of ΔH1 and ΔH2.
Let's consider the extreme case by assuming that ΔH3 is equal to the sum of ΔH1 and ΔH2 but with an opposite sign:
ΔH3 = -(ΔH1 + ΔH2)
ΔH3 = -(138 kJ + 241 kJ)
ΔH3 = -379 kJ
Therefore, the minimum value of ΔH3 that would make the overall reaction exothermic is -379 kJ. If ΔH3 is equal to or less than -379 kJ, the overall ΔH value will be negative, indicating an exothermic reaction.
In summary, the minimum value of ΔH3 that would make the overall reaction exothermic is -379 kJ, assuming the values of ΔH1 and ΔH2 remain the same.