S8 (rhombic) + 12 O2 (g)-----> 8 SO3 (g) + 752.0 kcal
In the reaction above equated, performed at 25 °c and 1 atm, the formation enthalpy of the SO3 (g) is equal to:
a) +752,0 kcal
b) -752,0 kcal
c) +773,0 kcal
d) -94,0 kcal
e) +94,0 kcal
The reaction is exothermic of 752.0 kJ so dHf = -752.0 kJ/8 mols SO3 or -752.0/8 kJ/mol SO3
thank u :)
To find the formation enthalpy of SO3 (g) in the given reaction, we need to use the concept of Hess's Law and standard enthalpies of formation.
Hess's Law states that the enthalpy change in a chemical reaction is independent of the pathway taken. It means that we can calculate the overall enthalpy change of a reaction by considering the enthalpy changes of the individual steps required to reach the desired reaction.
The standard enthalpy of formation (ΔHf°) is defined as the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states, with all reactants and products in their standard states.
In this case, we need to consider the standard enthalpies of formation of S8 (rhombic), O2 (g), and SO3 (g) to calculate the formation enthalpy of SO3 (g).
The standard enthalpy of formation of an element in its standard state is defined as zero. Therefore, the standard enthalpy of formation of S8 (rhombic) and O2 (g) is zero.
Now let's consider the reaction and calculate the formation enthalpy of SO3 (g):
S8 (rhombic) + 12 O2 (g) → 8 SO3 (g)
Since the standard enthalpy of formation of S8 (rhombic) and O2 (g) is zero, the overall enthalpy change in the reaction is equal to the formation enthalpy of SO3 (g).
Given: 8 SO3 (g) + 752.0 kcal
Therefore, the formation enthalpy of SO3 (g) is +752.0 kcal (option a).