Posted by Annabelle on .
change in H(rxn)= -296.06 kJ/mol
change in H(rxn)= -296.36 kJ/mol
1. Calculate the enthalpy change for the transformation. S(rhombic)-->S(monoclinic)
Our goal is to calculate the enthalpy change for the formation of monoclinic sulfur from the rhombic sulfur. To do soo, we must arrange the equations that are given in the problem in such a way that they will sum to the desired overall equation.
So which equation should be reversed: 1st or 2nd?
What is the sign of change in H for the reversed equation: positive or negative?
Chemistry- please help ASAP!!!! thank you -
This is an old question, but I'm sure someone could benefit by this answer.
You must reverse the 1st equation because you want S(rhombic) on the left and S(monoclinic) on the right in the end equation:
S(rhombic) --> S(monoclinic)
When you reverse a reaction, delta H gets the opposite sign. In this case, delta H = 296.36 kJ/mol (now positive) for the first equation when flipped. Now you can cancel like terms in the equation to get S(r) ==> S(m), and add the two delta H's together to get delta H for this new reaction.