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?
Use equation 1 as is.
Reverse equation 2 (changer sign, too) and add to equations. Add delta Hs.
Cancel ions common to each side. You should have
S(rhombic) ==> S(monoclinic)
Thank you so much
In order 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.
Let's examine the two given 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
To obtain the desired overall equation S(rhombic) --> S(monoclinic), we need to reverse either the 1st or 2nd equation.
Let's analyze the options:
- If we reverse the 1st equation, it becomes SO2(g) --> S(rhombic) + O2(g), but this is not the desired overall equation.
- If we reverse the 2nd equation, it becomes SO2(g) --> S(monoclinic) + O2(g), which matches the desired overall equation.
Therefore, we should reverse the 2nd equation.
Now, let's determine the sign of the change in enthalpy for the reversed equation SO2(g) --> S(monoclinic) + O2(g).
Since the given change in enthalpy for the original equation S(monoclinic) + O2(g) --> SO2(g) is -296.36 kJ/mol, the sign of the change in enthalpy for the reversed equation will be the opposite.
Therefore, the sign of the change in enthalpy for the reversed equation SO2(g) --> S(monoclinic) + O2(g) is positive.
To calculate the enthalpy change for the transformation S(rhombic) to S(monoclinic), we need to rearrange the given equations.
The given equations are:
1) S(rhombic) + O2(g) ---> SO2(g) (change in H = -296.06 kJ/mol)
2) S(monoclinic) + O2(g) ---> SO2(g) (change in H = -296.36 kJ/mol)
We want the overall equation: S(rhombic) ---> S(monoclinic)
To achieve this, we need to reverse one of the equations. Let's analyze the sign of change in H.
For the forward reaction in the first equation (S(rhombic) + O2(g) ---> SO2(g)), the change in H is negative (-296.06 kJ/mol).
Now, when we reverse this equation, it becomes SO2(g) ---> S(rhombic) + O2(g).
For the reverse reaction, the sign of change in H is opposite to that of the forward reaction. So, the sign of change in H for the reversed equation is positive.
Therefore, we need to reverse the first equation (S(rhombic) + O2(g) ---> SO2(g)), and the sign of change in H for the reversed equation is positive.