Sure! To solve for the adjusted enthalpy change for Reaction 2 when it needs to be tripled, we can follow Hess' Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.
Given the enthalpy change for the original Reaction 2:
[tex]\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O , \Delta H = -286 \, \text{kJ} \][/tex]
We need to multiply this reaction by 3, as indicated. Let’s denote the multiplier by 3. Therefore, the reaction becomes:
[tex]\[ 3 \times \left( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \right) , \Delta H_{\text{adjusted}} \][/tex]
When we multiply the reaction coefficients by 3, we also need to multiply the enthalpy change by 3 to maintain consistency:
[tex]\[ \Delta H_{\text{adjusted}} = 3 \times (-286 \, \text{kJ}) \][/tex]
Therefore, the adjusted enthalpy change for the reaction is:
[tex]\[ \Delta H_{\text{adjusted}} = -858 \, \text{kJ} \][/tex]
So, the correct adjusted enthalpy for Reaction 2 when it is tripled would be:
[tex]\[ \boxed{-858 \, \text{kJ}} \][/tex]
