Which element is a noble gas? A. H B. Ko C. Xx D. Ar E. O
2 answers:
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The question is incomplete, here is the complete question.
Consider these reactions, where M represents a generic metal.
1. ; ΔH = -783 kJ
2. ; ΔH = -74.8 kJ
3. ; ΔH = -1845 kJ
4. ; ΔH = -195 kJ
Use the information above to determine the enthalpy of the following reaction.
; ΔH=?
Answer : The enthalpy for the reaction is, -6376.8 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
Now we have to determine the enthalpy of the reaction.
Multiplying reaction 2 by 6, reaction 3 by 3 and reverse reaction 4 by 2 then adding all the reaction, we get:
1. ; ΔH = -783 kJ
2. ; ΔH = 6 × (-74.8 kJ) = -448.8 kJ
3. ; ΔH = 3 × (-1845 kJ) = -5535 kJ
4. ; ΔH = 2 × 195 kJ = 390 kJ
The expression for enthalpy for the reaction is,
Therefore, the enthalpy for the reaction is, -6376.8 kJ