Phosphorus pentachloride forms from phosphorus trichloride and chlorine:
(a) Use data in Appendix B to calculate ∆S_sys, ∆S_surr, and ∆S_total for this reaction. Is the reaction spontaneous under standard-state conditions at 25 °C?

Question

Phosphorus pentachloride forms from phosphorus trichloride and chlorine:

(a) Use data in Appendix B to calculate ∆S_sys, ∆S_surr, and ∆S_total for this reaction. Is the reaction spontaneous under standard-state conditions at 25 °C?

Accepted Answer

well, everyone in this video are being asked to calculate for the delta sf system, delta sf surroundings and the delta S total for the reaction. And we're talking about this reaction right on top. So we're also being asked if this reaction is going to be spontaneous or non spontaneous. So let's go ahead and recall that my delta S total. If this is greater than zero. So if we have a positive value, the reaction is going to be spontaneous, but if our delta as total is less than or equal to zero, this reaction is going to be non spontaneous. So first we can go ahead and calculate for my delta as total. So my delta as total. This has a formula equal to delta S. Of system dusty delta S of surroundings, Calculate for the Delta S. of the system is also equal to the Delta S0 which is equal to the entropy of my products minus the entropy of my reactant. All right. And we know that my delta S of surroundings has a formula that's equal to negative delta H over T for temperature. And within this equation my delta H is equal to the delta age. My products minus the delta H. Of my react ints. So they are just tv. All of our formulas that we have. And we can go ahead and calculate for each thing that we need to calculate for this problem. Starting off with my delta S of the system. This has the formula we're gonna go ahead and use the formula of the entropy or the entropy of my products minus the entropy of the reactant. So we have the Delta S. of the system equaling to, well we have one mole of oxygen gas multiplying this by 205 jewels per kelvin times mold. And this value is just from the charts here. So in our case this right here and we're using this one mole here to go ahead and counsel our mole units and leave us with just jewels per Calvin's. Alright so continue on. And then we're adding this with one mole of silicon Multiplied by 18.8 jewels per Calvin times more. And we're subtracting this with one mole of S. I. O. To multiplying this with 41. jewels per kelvin times more. And once you put everything into calculator we see that we get the delta S. Of the system to equal 282 jewels per Calvin's. Alright so this is one part of our answer for this question. Now next what we're gonna do is we can go ahead and solve or calculate for our adult to H. So my delta H. Value is equal to again we have to do some cancelations for the mole unit here. So we have one mole of 02 multiplied by zero killer jewels per mole plus one mole of silicon multiplied with zero killer jewels per mole And subtracting this with the one mole of S. I. 02 multiplied by negative 910.7 killer jewels per mole. So what you see here is that this automatically all zero. So is this right here and then the right side for this one, we can see that my units of mole will go ahead and cancel. So we'll just be left with the final answer to have the units of being killed jules. So once I put everything into my calculator, I see that my delta H value is an equal to 910.7 kg jewels. We want this actually to be in jewels so we can go ahead and do a direct conversion here. So for every 1000 jewels we have one killer jewel, the units of cultural council here, we can get my final dr H value in terms of jewels to be 9.107 times 10 to the fifth power. Now, what we can do is change our Celsius temperature into kelvin's. So for our temperature We're giving this to be 25°C. We can go ahead and add 273. to this value. To convert our degreeC into Calvin's. So this gives me a value of 298.15 calvins. Now we're gonna go ahead and calculate then for my delta s of surroundings. So this equals to negative 9.107 times 10 to the fifth jewels. We're going to divide this with my tea. So that's 298.15 kelvin's. Whereas using our written formula above all right. So the delta as of surroundings. When I put everything into the calculator, I get the value to be negative 3.5 times 10 to the third. Power units being jewels per kelvin. So this is my delta S. Of surroundings value. Now moving on to my delta as total. So now that I have all the components of the delta S total, which is the delta S. System plus the delta sf surroundings which go ahead and plug in those values. So that's negative 3054.5 joules per Calvin plus my 182 jewels per kelvin value was I put that into a calculator. My sum is equal to negative jules per kelvin. Gonna write this in the correct amount of significant figures and that's three to be exact. So go ahead. Also use scientific notation, so it's equal to negative 2.87 times 10 to the third jewels per kelvin value. So we can see here that my delta as total is definitely less than or equal to zero. Since we have a negative value. This means then that the reaction is non spontaneous. And this answer is that if the reaction is going to be spontaneous or not. This reaction is not going to be spontaneous

Phosphorus pentachloride forms from phosphorus trichloride and chlorine:
(a) Use data in Appendix B to calculate ∆S_sys, ∆S_surr, and ∆S_total for this reaction. Is the reaction spontaneous under standard-state conditions at 25 °C?

Verified Solution

Key Concepts

Entropy (∆S)

Gibbs Free Energy (∆G)

Standard-State Conditions

Phosphorus pentachloride forms from phosphorus trichloride and chlorine:(a) Use data in Appendix B to calculate ∆Ssys, ∆Ssurr, and ∆Stotal for this reaction. Is the reaction spontaneous under standard-state conditions at 25 °C?

Verified Solution

Key Concepts

Entropy (∆S)

Gibbs Free Energy (∆G)

Standard-State Conditions

Phosphorus pentachloride forms from phosphorus trichloride and chlorine:
(a) Use data in Appendix B to calculate ∆S_sys, ∆S_surr, and ∆S_total for this reaction. Is the reaction spontaneous under standard-state conditions at 25 °C?