Sulfur trioxide is formed by the catalytic oxidation of sulfur dioxide:

2SO2(g)+O2(g)→2SO3(g)
If ΔH∘= -197.8kJ/mol and ΔS∘= -188.0J/K , what is ΔStotal for this reaction?
Is the reaction spontaneous under standard-state conditions at 25 ∘C?

dH surroundings = -dH/T

dStotal = dH surroundings + dH system (in the problem)

Awesome thank you!

How would I determine when the reaction goes nonspontaneous?

If dStotal is +, the reaction is spontaneous.

To determine the total change in entropy (ΔStotal) for the reaction, you can use the formula:

ΔStotal = ΔS(products) - ΔS(reactants)

First, let's calculate the change in entropy for the products (SO3):

ΔS(products) = n(SO3) × ΔS°(SO3)

Since there are 2 moles of SO3 formed in the reaction, we multiply the ΔS° value by 2:

ΔS(products) = 2 × ΔS°(SO3)

Next, let's calculate the change in entropy for the reactants (SO2 and O2):

ΔS(reactants) = n(SO2) × ΔS°(SO2) + n(O2) × ΔS°(O2)

In the balanced equation, we see there are 2 moles of SO2 and 1 mole of O2:

ΔS(reactants) = 2 × ΔS°(SO2) + 1 × ΔS°(O2)

Now, substitute the given values for ΔS° into the equation:

ΔS°(SO3) = -188.0 J/K

ΔS°(SO2) = 0 J/K (since it is a pure element)

ΔS°(O2) = 0 J/K (since it is a pure element)

Plugging these values into the equation, we have:

ΔS(products) = 2 × (-188.0 J/K) = -376 J/K

ΔS(reactants) = 2 × 0 J/K + 1 × 0 J/K = 0 J/K

Finally, calculate ΔStotal:

ΔStotal = ΔS(products) - ΔS(reactants) = (-376 J/K) - 0 J/K = -376 J/K

Now, to determine if the reaction is spontaneous under standard-state conditions at 25 ∘C, you can use the Gibbs free energy equation:

ΔG° = ΔH° - TΔS°

where ΔH° is the enthalpy change and ΔS° is the entropy change at standard conditions (STP).

Given that ΔH° is -197.8 kJ/mol and ΔS° is -188.0 J/K, we need to convert them to J before plugging them into the equation:

ΔH° = -197.8 kJ/mol × (1000 J/1 kJ) = -197800 J/mol

ΔS° = -188.0 J/K

Plugging these values into the equation:

ΔG° = (-197800 J/mol) - (298 K) × (-188.0 J/K) = -197800 J/mol + 56104 J/mol = -141696 J/mol

Since ΔG° is negative, the reaction is spontaneous under standard-state conditions at 25 ∘C.