S(rhombic)+O2(g)--->SO2(g)

change in H(rxn)= -296.06 kJ/mol

S(monoclinic)+O2(g)--->SO2(g)
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?

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.

Well, if we want to calculate the enthalpy change for the transformation of S(rhombic) to S(monoclinic), we need to reverse one of the given equations.

To determine which equation to reverse, we compare the reactants and products of the desired transformation (S(rhombic) to S(monoclinic)) with the reactants and products of the given equations.

Looking at the given equations:
1. S(rhombic) + O2(g) ---> SO2(g) with ∆H = -296.06 kJ/mol
2. S(monoclinic) + O2(g) ---> SO2(g) with ∆H = -296.36 kJ/mol

We see that both equations have SO2 as a product, but only the second equation has S(monoclinic) as a reactant. Therefore, we need to reverse the second equation.

Now, as for the sign of the change in enthalpy (∆H) for the reversed equation, it will be the opposite sign of the original equation. So, if ∆H for the second equation is negative (-), then the ∆H for the reversed equation will be positive (+).

To calculate the enthalpy change for the transformation S(rhombic) --> S(monoclinic), we need to reverse one of the given equations and determine the sign of the change in enthalpy for the reversed equation.

To decide which equation to reverse, we should consider the desired reaction: S(rhombic) --> S(monoclinic). Looking at the given equations, we can see that the sulfur on the reactant side of both equations is in the same form as the sulfur on the product side of the desired reaction. Therefore, we need to reverse the equation where sulfur is in a different form.

In this case, the sulfur is in rhombic form in the first equation (S(rhombic) + O2(g) --> SO2(g)), and it is in monoclinic form in the second equation (S(monoclinic) + O2(g) --> SO2(g)). Therefore, we need to reverse the first equation.

Now, we need to determine the sign of the change in enthalpy for the reversed equation. Reversing an equation changes the sign of the change in enthalpy. Since the given change in enthalpy for the first equation is -296.06 kJ/mol, reversing the equation will result in a change in the sign of the enthalpy value.

Thus, the sign of the change in enthalpy for the reversed equation is positive (+).

To calculate the enthalpy change for the transformation S(rhombic) --> S(monoclinic), we need to rearrange the given equations.

First, let's compare the two equations:

1) S(rhombic) + O2(g) ---> SO2(g) ΔH = -296.06 kJ/mol
2) S(monoclinic) + O2(g) ---> SO2(g) ΔH = -296.36 kJ/mol

Since we want to go from rhombic sulfur to monoclinic sulfur, we need to reverse one of the equations. This will make the sulfur on the left side of the equation and on the right side of the equation match.

Let's reverse equation 2):

Reverse of equation 2: SO2(g) ---> S(monoclinic) + O2(g)

Now, let's check the sign of the enthalpy change for the reversed equation. When we reverse an equation, the sign of the enthalpy change also changes.

The enthalpy change for equation 2 is -296.36 kJ/mol. Therefore, the enthalpy change for the reversed equation is +296.36 kJ/mol.

So, to summarize:
Reversed equation: SO2(g) ---> S(monoclinic) + O2(g)
Enthalpy change for reversed equation: +296.36 kJ/mol