Calculate ∆G° for the following reaction: P 4 (s) + 5 O 2 (g) → P 4 O 10 (s), ∆H° = −2940 kJ/mol, 25 °C. Does the reaction favor reactants or products?

Here it says calculate the change in the standard Gibbs free energy or the following reaction, where we have p 4 solid reacting with 5 moles of oxygen to give us 1 mole of tetraphosphorus dioxide. Here they tell us that our change in the enthalpy, standard enthalpy, is negative 29 40 kilojoules per mole at 25 degrees Celsius. Does the reaction favor reactants or products? Alright. So we're going to say here that they're giving us the entropies of formation. The entropies of formation here will help me determine the entropy of my reaction. Remember that's equal to products minus reactants. So So that'll equal we have 1 mole of P4010 product, so that's 1 mole of P4010, with an entropy of formation of 228.9 joules over moles times k, minus your reactants where we have 1 mole of phosphorus solid, which is 164.4 joules over moles times k, plus we have 5 moles of 2, each one is 205.2 joules over moles times k. Here we'd say that our moles cancel out, so we'll have the entropy of our reaction equal to joules over Kelvin, so that comes out to negative 961.5 joules per Kelvin. Alright. Now we're going to say, we know what delta s is, we know what delta h is, we know what our temperature is. We can plug all these values in and solve. So here we're going to say that delta G equals delta H minus T delta S. Here we're gonna plug in our delta h, so here this is kilojoules per mole, so this is just negative 9240, kilojoules minus temperature. You add 273.15 to the degree Celsius to give us 298.15 Kelvin. We change the delta s that we calculated into in kilojoules by dividing it by a 1000, so that's 0.9615 kilojoules per Kelvin. Here, Kelvins cancel out, so we'll have our answer at the end in kilojoules. This comes out to negative 2653 kilojoules as our final answer. So this would represent the change in our standard Gibbs free energy.

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19. Chemical Thermodynamics

Gibbs Free Energy Calculations

General Chemistry

19. Chemical Thermodynamics

Gibbs Free Energy Calculations

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Multiple Choice

Calculate ∆G° for the following reaction: P₄ (s) + 5 O₂ (g) → P₄O₁₀ (s), ∆H° = −2940 kJ/mol, 25 °C.
Does the reaction favor reactants or products?

1419.3 kJ

−2653 kJ

−140.7 kJ

598.5 kJ

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Verified step by step guidance

Identify the given values: ∆H° = −2940 kJ/mol, temperature T = 25 °C (which is 298 K), and the standard entropy values from the image: S°(P4) = 41.1 J/mol•K, S°(O2) = 205.0 J/mol•K, S°(P4O10) = 340.0 J/mol•K.

Calculate the change in entropy (∆S°) for the reaction using the formula: ∆S° = ΣS°(products) - ΣS°(reactants). Convert the entropy values from J/mol•K to kJ/mol•K by dividing by 1000.

Apply the formula for Gibbs free energy change: ∆G° = ∆H° - T∆S°. Use the calculated ∆S° and the given ∆H° and temperature T to find ∆G°.

Interpret the sign of ∆G°: If ∆G° is negative, the reaction favors the formation of products. If ∆G° is positive, the reaction favors the reactants.

Compare the calculated ∆G° with the given options to determine which value corresponds to the correct answer.

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Struggling with General Chemistry?

Calculate ∆G° for the following reaction: P4 (s) + 5 O2 (g) → P4O10 (s), ∆H° = −2940 kJ/mol, 25 °C.Does the reaction favor reactants or products?

Watch next

Gibbs Free Energy Calculations practice set

Calculate ∆G° for the following reaction: P₄ (s) + 5 O₂ (g) → P₄O₁₀ (s), ∆H° = −2940 kJ/mol, 25 °C.
Does the reaction favor reactants or products?