Phosphorus pentachloride forms from phosphorus trichloride and chlorine: PCl 3 (g) + Cl 2 (g) → PCl 5 (g) (b) Estimate the temperature at which the reaction will become nonspontaneous.

Hello. Everyone in this video being asked at one temperature does the formation of silicon oxide from the reaction of solid silicone and oxygen gas become non spontaneous? Let's recall that went down to us total value is greater than zero, we have a spontaneous reaction. So basically when our delta S total value is greater than zero which is positive, we have a positive delta S. Value. Then this will be spontaneous. But if my delta S of total is less than or equal to zero and this reaction will be non spontaneous. Alright so let's go ahead and calculate for the delta S total then. So we have a formula for this. We know that delta S total is equal to delta us of the system plus the delta us of the surroundings. So we can plug in those values. So we have that might delta S. Of the system is also equal to delta S. Zero which is equal to the and the P. Or the entropy of our product minus the entropy of the reactant. Alright, so now solving for my tell to us of the system here. So we have one more of our silicon oxide. So that S. I. 02. And then we're gonna go ahead and multiply this with four or 41. joules per kelvin times mall. And then we're gonna go ahead and subtract this With one more of Silicon, multiplied by 18. Jewels for Calvin Times Mall and then adding this With one mole of oxygen gas, multiplied by 205.0 joules per kelvin times mold. Alright, so once you put everything in to the calculator. And you see that my delta S. Of the system is equal to negative 82.3 jewels per kelvin's. So let's go ahead and calculate for the delta S. Of the surroundings. Go go ahead and scroll down for more space here. So for my formula for the delta S. Of the surroundings is equal to negative DELTA H. Over my temperature. So we know that the doctor each is equal to the adult age of formation of the product minus the delta age, a formation of my reactant. All right, so doing the same thing here as we did earlier, we have one more of S. I. O. To multiply this by negative 910.7. Kill jules per mole subtracting this with what we have one more of silicon notified by zero plus one mole of + multiplied by zero. What we what this gives us is that the delta H. Is equal to -9 10.7. Gonna convert this into jewels sell for every 1000. So again we have this one kg jewels. So very 1000 jewels. We have one kg jewel. This gives us that my delta H value is equal to negative 9.107 times 10 to the five jewels. So now finally we can go ahead and solve for the temperature. So t when my delta as a total is equal to zero, which we can find the temperature for when this reaction is non spontaneous. So again, zero is what the s total is. And this is equal to the delta S of the system plus the delta s. Of the surroundings, which we have the values of both. So we go ahead and just plug those numbers in from my delta S of system, this is negative 182.3 units being jewels for Calvin. And then for my delta s of surroundings, gonna go at this with a negative negative 9.107 times 10 to the fifth power of jewels over R. T. Value here. Doing some simplification by moving our first number here, over to the left of the equation sign, get a positive 182.3 jewels per kelvin equaling to a positive 9. times 10 to the fifth jewels over. T for temperature. Now we go ahead isolate T. And we get 900 or 9.107 times 10 to the fifth power of jewels Over my 182.3 jewels per kelvin. We'll see now for our units saying that the jewels units will cancel leaving us with the units of temperature just to be Calvin's which is correct. So once I put these numerical values into the calculator, I get the final value of 4995.66 or 611 kelvin's one around this to three significant figures, and we can go ahead and do this by using scientific notation. So T for my final answer is equal to 5.0 times 10 to the third power kelvin's. So this right here is going to be my final answer for the problem.

Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

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McMurry 8th Edition

Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

Problem 77b

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Chapter 18, Problem 77b

Phosphorus pentachloride forms from phosphorus trichloride and chlorine: PCl₃(g) + Cl₂(g) → PCl₅(g) (b) Estimate the temperature at which the reaction will become nonspontaneous.

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Determine the standard enthalpy change (\( \Delta H^\circ \)) for the reaction using standard enthalpies of formation.

Determine the standard entropy change (\( \Delta S^\circ \)) for the reaction using standard molar entropies.

Use the Gibbs free energy equation \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \) to find the temperature at which \( \Delta G^\circ = 0 \), indicating the reaction is at equilibrium and thus nonspontaneous.

Rearrange the equation to solve for temperature: \( T = \frac{\Delta H^\circ}{\Delta S^\circ} \).

Substitute the values of \( \Delta H^\circ \) and \( \Delta S^\circ \) into the equation to estimate the temperature.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy

Gibbs Free Energy (G) is a thermodynamic potential that helps predict the spontaneity of a reaction. A reaction is spontaneous at constant temperature and pressure if the change in Gibbs Free Energy (ΔG) is negative. The relationship between ΔG, enthalpy (ΔH), and entropy (ΔS) is given by the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.

Entropy (S)

Entropy is a measure of the disorder or randomness in a system. In chemical reactions, an increase in entropy typically favors spontaneity. For the reaction of phosphorus trichloride and chlorine to form phosphorus pentachloride, understanding how the number of gas molecules changes can help assess the change in entropy and its impact on the reaction's spontaneity.

Enthalpy (ΔH)

Enthalpy is a measure of the total heat content of a system and is crucial for understanding energy changes during a reaction. The change in enthalpy (ΔH) can be either exothermic (releasing heat) or endothermic (absorbing heat). For the reaction in question, knowing whether it is exothermic or endothermic will help determine the temperature at which the reaction becomes nonspontaneous, as it influences the Gibbs Free Energy equation.