calculate the enthalpy change for the allotropic transformation of sulphur in the monoclinic form to sulphur in the rhombic form from the following data's
1.S+oxygen gives sulphurdioxide,Delta H is equal to-295.1KJ
2.sulphur +oxygen gives sulpherdioxide,Delta H is equal to -296.KJ
To calculate the enthalpy change for the allotropic transformation of sulphur in the monoclinic form to sulphur in the rhombic form, we can use the Hess's Law of Additivity of Enthalpies.
The allotropic transformation can be divided into two steps:
1. Monoclinic Sulphur to Sulphur Dioxide:
- Equation: S (monoclinic) + O2 -> SO2
- Enthalpy Change (ΔH1): -295.1 KJ
2. Sulphur Dioxide to Rhombic Sulphur:
- Equation: SO2 -> S (rhombic) + O2
- Enthalpy Change (ΔH2): -(-296 KJ) = 296 KJ (change in sign as we are going in the reverse direction)
Now, to calculate the enthalpy change for the overall transformation from monoclinic sulphur to rhombic sulphur, we can sum up the ΔH values:
ΔH_total = ΔH1 + ΔH2
= -295.1 KJ + 296 KJ
= 0.9 KJ
Therefore, the enthalpy change for the allotropic transformation of sulphur in the monoclinic form to sulphur in the rhombic form is 0.9 KJ.
To calculate the enthalpy change for the allotropic transformation of sulphur from monoclinic to rhombic form, we will need to use the principle of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
In this case, we can consider the reaction pathway as follows:
1. Monoclinic Sulfur -> Sulfur Dioxide:
This reaction is given as S + Oxygen -> Sulfur Dioxide, with ΔH equal to -295.1 kJ.
2. Rhombic Sulfur -> Sulfur Dioxide:
This reaction is given as S + Oxygen -> Sulfur Dioxide, with ΔH equal to -296.0 kJ.
Now, since we want to calculate the enthalpy change for the allotropic transformation of sulfur, we can see that both reactions involve the formation of sulfur dioxide. However, the initial and final forms of sulfur differ.
To calculate the enthalpy change for the transformation, we need to flip and scale one of the reactions to match the stoichiometry of the other reaction.
Let's reverse the second reaction and multiply it by 1/2:
1/2 * (Sulfur Dioxide -> Rhombic Sulfur):
This reaction is given as 1/2 Sulfur Dioxide -> S, with ΔH equal to +148.0 kJ (since we reversed the sign).
Now, we can add up the enthalpy changes of the reactions to get the enthalpy change for the allotropic transformation:
ΔH(allotropic transformation) = ΔH(monoclinic sulfur -> Sulfur Dioxide) + ΔH(1/2 * Sulfur Dioxide -> Rhombic Sulfur)
ΔH(allotropic transformation) = (-295.1 kJ) + (+148.0 kJ) = -147.1 kJ.
Therefore, the enthalpy change for the allotropic transformation of sulfur from monoclinic to rhombic form is -147.1 kJ.
Note: Data is the plural of datum. Apostrophe not needed.
Let Sr = sulfur rhombic and Sm = sulfur monoclinic.
You don't specify which is which; however, check your data.
I found Sr + O2 ==> SO2(g) dH = -296.06
......... Sm + O2 --> SO2(g) dH = -296.36.
If we assume your data to be mean #1 is monoclinic
.........Sm + O2 ==> SO2(g) dH = -295.1 and #2 to be rhombic so
........Sr + O2 ==> SO2(g) dH = -296.
Take equation 1 and add to the reverse of equation 2 like this.
.............Sm + O2 ==> SO2(g) dH = -295.1
.............SO2(g) ==> Sr + O2 dH = +296
------------------------------------------------------------
...........Sm ==> Sr -295.1 + 296 = +?
If I have assumed incorrectly you can easily change the two equations and get your answer. Again, please check your numbers.