Using the data in Appendix C and given the pressures listed, calc...

Question

Using the data in Appendix C and given the pressures listed, calculate Kp and ΔG for each of the following reactions:
(c) N2H41g2¡N21g2 + 2 H21g2 PN2H4 = 0.5 atm, PN2 = 1.5 atm, PH2 = 2.5 atm

Accepted Answer

Welcome back everyone in this example, we have the partial pressures for nitrogen phosphorus ammonia and tetra phosphorous given above here were also given the standard gibbs free energy of formation of each of these species as well. And we need to calculate the equilibrium constant for partial pressure and change in gibbs free energy at 2 98 kelvin for the below reaction which is a balanced equilibrium reaction that we're given here because we have an equilibrium arrow. So what we're going to begin with is finding the standard change in gibbs free energy of our reaction, delta G. Degree of our reaction. And recall that we can calculate this by taking the standard change and gives free energy of formation of our products subtracted from our standard change and gibbs free energy of formation of our reactant and will represent our reactant in the color red. So let's go ahead and plug in what we know because we are given all the gifts. Free energy change of formations for each of our reaction species. So we would say that delta G. Of our reaction is equal to for the some of the gives free energy of formation of our products. So sorry, we need to include that some symbol there in our formula we're going to begin with our Gibbs. Free energy of formation of tetra phosphorous given as 24.4 kg joules per mole. And according to our formula, we only have one mole of tetra phosphorous. So let's actually include that. So everything's clear. One more. Sorry, one mole of tetra phosphorous multiplied by its Gibbs free energy of formation of 24.4. This is then added to our next product which is our formals of Ammonia, multiplied by ammonia as gibbs free energy of formation being a value of negative 16.4 kg joules per mole. As given in the prompt. So that completes our gibbs free energy of formation of our products. So we're gonna end our bracket off here and we want to subtract from the gibbs. Free energy of formation of our reactant. So beginning our brackets we have for our first reactant, one mole of our and sorry, not one mole, but four moles. Because we see we have phosphene formals of that. Four moles of phosphate in P H three multiplied by phosphenes. Standard gibbs free energy of formation which is a value given in the prompt of 13.5 kg joules per mole. This is then added to our second reactant where we have two moles of our nitrogen gas, multiplied by its standard- Gibbs free energy of formation which given in the prompt is the value of zero kg Permal. And this completes the some of the standard gibbs free energy of formation of our reactant and products. And now we can take the difference between these values but we can just simplify this just so it's clear. So we would have 24.4 kg joules per mole. And actually let's just do everything in one step. So in our calculators, we're gonna type all of this in for our first bracket. That's going to give us a value of -41.2 kg joules per mole. Which is then subtracted from our some of our change. And gibbs free energy of our formation of our reactant. Which would give us a value of 54 kg joules per mole. And this will simplify to a value of negative 95.2 kg jules Permal as our standard gibbs free energy change of our reaction. And now that we know this value, we can use it to find our change in gibbs free energy at 2 98 kelvin for the reaction. By recalling that we would utilize the following formula where we would say it's calculated from taking our change in standard gibbs free energy added to our gas constant R. Which is multiplied by our temperature in kelvin, which is multiplied by the Ln of our reaction quotient. Q. And actually, let's use the color blue for our standard gibbs Free energy since we calculated it above. So now that we know this equation, we want to recognize what our reaction quotient is for Q. And so we want to take note of our partial pressures given in the prompt of each of our reactant and products recall that our reaction quotient represents our partial pressure of our products raised to their coefficient divided by the partial pressure of our reactant raised to their coefficients. And so we want to make note of our partial pressures given in the prompt. So we're given the partial pressure. First of our reactant, nitrogen gas equal to our partial pressure of our phosphene gas as the value 0.6 A. T. M's. Were given our partial pressure of our product ammonia as the value 0.3 A. T. M. And were given the partial pressure of our product, touch or phosphorus as point to A. T. M. So plugging in what we know for our reaction quotient, we can say that we have for our partial pressure of our products beginning with our partial pressure of nitrogen as well as its coefficient as now an exponents to two because we have two moles of our nitrogen which is multiplied by our second product which is our phosphene where we have our partial pressure of ph three phosphene raised to its coefficient. Where we have formals of our phosphene in our denominator, we plug in our partial pressure of our reactant where we have our ammonia ammonia. And in our equation we counted four moles of ammonia. So that's raised to an exponent of four, multiplied by the partial pressure of our tetra phosphorous which is raised to an exponents of just one, which we don't have to write in since we only have one mole of it. And actually I just realized that this is backwards because ammonia and tetra phosphorous, our reactant and phosphorus and nitrogen are products. So this should be in the denominator. And this should be in the numerator here. So plugging in our partial pressure values, we would have our partial pressure of ammonia given in the prompt as 0.3 A. T. M's raised to the exponent four, multiplied by our numerator. Where we have our partial pressure of tetra phosphorous as point to A T. M's given in the prompt where we just have one mole of that. In our denominator we have our partial pressure of nitrogen equal to our partial pressure of phosphene as 0. A. T. M's First raised to the second power for nitrogen and then 0.0. This is multiplied by 0.06 ATMs for a partial pressure of phosphene raised to an exponent of four because they both have the same partial pressure. And so in our calculators we're going to get that our reaction quotient is equal to a value of 34,722. All of our units of a TMS will cancel out. And now that we know our value for Q. We can calculate our change in gibbs free energy set that equal to our standard gibbs. Free energy change which we calculated above as negative 95.2 kg joules per mole which we add to our gas constant. R which we should recall is the value 8.314 with units of joules per moles times kelvin. This is then multiplied by our temperature in kelvin. Given in the prompt as 298 kelvin. And then we have the L. N. Of our reaction quotient also multiplied by this. So the Ln of our reaction quotient that we solved for three or 34,722. And this will give us our change in gibbs free energy equal to canceling out our units. First. We can get rid of our units of kelvin with kelvin here we can get rid of our units of jewels by expanding our answer for our standard gibbs. Free energy and converting it from kilo jewels, two jewels. So we're gonna actually adding a step here and multiply by the conversion factor to go from kilo jewels. Two jewels by recalling that are prefix kilo, tells us that for one kg jule we have 10 to the third power jewels. And this allows us to cancel out kilo jewels as well as jewels. Actually sorry we are left with jewels because they're both in the numerator. So carefully plugging this into our calculators were going to yield a result negative 69,296 jewels Permal which we will recall that are changing gibbs. Free energy should still be in units of kilo joules per mole. So we're going to go ahead and just multiply by the conversion factor to go from jules back to kill a jewels by recalling that are prefix kilo tells us that we have for one kg jule tend to the third power jewels, allowing us to cancel out our units of jewels. Leaving us with kilo joules per mole. And this is going to give us a value of negative 69. kg jules. I'm sorry, this is kilo joules per mole. So kj per mole. So now we have one of our first answers here as our change in gibbs free energy of our reaction. And now we want to continue our solution to calculate our equilibrium constant of partial pressures. So this is to find K. P equilibrium constant of partial pressure. Where we want to recall the formula where we can take our standard gibbs free energy change set equal to negative one times r. Gas constant R. Which is then multiplied by the temperature in kelvin, which is then multiplied by the Ln of our equilibrium constant. K. So we'll use the color purple for that. And so to isolate for our equilibrium constant K. We can say that K. And actually just to show the work for that, we want to divide both sides by negative our times temperature. So we have this which is going to lead us into our standard gibbs free energy divided by negative our times our temperature in kelvin set equal to the natural log of our equilibrium partial pressure constant K. Where we would get rid of that Ln term by taking both sides as an exponents to the term E to cancel Ellen. And that would tell us that our equilibrium constant of partial pressures can be calculated from taking our standard gibbs free energy divided by I'm sorry. This is all to the power of, this is all as an exponents to the base of E. Where in our denominator we're dividing by our negative gas constant, multiplied by our temperature in kelvin. So at this point we just want to plug in all of our variables that we know. So we would say that k is equal to in our numerator, we should have our standard gibbs. Free energy change which we calculated above as -95.2 in our first step where we had units of kg per mole. And in our denominator we have our gas constant multiplied by negative one, which we should recall is negative or positive. 8.314 joules divided by moles times kelvin as our units multiplied by our temperature in kelvin Which is given in the prompt as 298 Kelvin. And this is all an exponent to the power or sorry, to the base of E. So we'll go ahead and make this clear by placing this all in brackets. Now recognize in our denominator we have again the term jewels meaning we want to convert our standard gibbs free energy from kilo jewels, two jewels so that jules can cancel out. So we're going to multiply our numerator by the conversion factor to go from kilo jewels. Two jewels by recalling that are prefix kilo tells us that we have 10 to the third power jewels, canceling out our units of kilo jewels. And also canceling out our units of jewels. We can also get rid of our terms moles as well as kelvin. So now we have no units which is what we want for our equilibrium partial pressure constant. And now we're going to carefully type this all into our calculators but actually just to show the work for that, I'll show you what the numerator and denominator should simplify too. So we would say that. And just to make more room one second, let's continue down for more room will say that our equilibrium constant of partial pressures is equal to In our numerator, we're going to have the simplification of negative 95,200 as our term in the numerator. And then in the denominator will have negative 24, 77.57. We also have no units because we canceled everything out. And so all we need to do is take this as an exponents to Eu alors number where we would say that our equilibrium constant should equal the value k equal to 4.8714 times 10 to the positive 16th power And we need to be careful with sig figs here. So to go back to this step where we have our quotient as an exponents. When we type this into our calculators, we should get a result of 38.42 47. And just to shorten that up we would just say 38.4, this is the result of our quotient and it has one decimal place. And so when we have Sig Figs to calculate for a constant and we have a race to a number with n decimal places and sorry, just so that's visible. We have to calculate for Sig Figs and we have users number rates to a power with n decimal places. The result should have n sync fix and because we stated we have one decimal place, that means therefore we have one Sig fig for our final answer. And so we would round this final answer to about one Sig fig as five times 10 to the 16th power as our equilibrium constant of partial pressures. So this would be our second final answer where our first answer again was our change in gibbs. Free energy for a reaction as we calculated to be negative 69.3 kg per mole and then be completed by finding our equilibrium constant of partial pressures as the value K equal to five times 10 to the 16 power with no units. This will complete this example as our final answer. I hope that everything I reviewed was clear. If you have any questions leave them down below and I'll see everyone in the next practice video.

Using the data in Appendix C and given the pressures listed, calculate Kp and ΔG for each of the following reactions: (c) N2H41g2¡N21g2 + 2 H21g2 PN2H4 = 0.5 atm, PN2 = 1.5 atm, PH2 = 2.5 atm

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