Use data in Appendix C to calculate ΔH°, ΔS°, and ΔG° at 25 °C fo...

Question

Use data in Appendix C to calculate ΔH°, ΔS°, and ΔG° at 25 °C for each of the following reactions.
(c) 2 P1s2 + 10 HF1g2¡2 PF51g2 + 5 H21g2

Accepted Answer

Hello everyone. So in this video we're going to go ahead and calculate for the change of entropy, the change of entropy and delta G a 25 degrees Celsius for this reaction right over here, of course we're giving all the values already in this table to go ahead and solve for each. So let's start off the problem by solving for our delta H. That's going to be that the delta H of my products is subtracted from the delta age of my reactions. So here we see that for our product side, we have two moles of a. L. C. L three and three moles of H two gas. So of course then we'll just go ahead and write that. So we have two moles of first, it's corresponding delta H value is negative 705.6. Let's see for the unit saying no be killer jewels Permal. For our second product we have the three moles of H two gas. The corresponding down to age value is zero kill jules per mole. So again, that's the delta age of my products. Let's go ahead and solve for the delta age of my reactant. So we have the first reactant having two moles. The corresponding delta H value is zero killer jewels per mole. Then for my second reactant we have six moles of the hydro hydrochloric acid. The corresponding delta H value is negative 92.30 killer jewels per mole. You can see here that the numerical values. Then if you put this into a calculator is going to be negative 857 .4 kg jewels. Then if you want to write this in this correct significant figures, we can go ahead and use scientific notation. So that's negative 8.57 times 10 to the second power and units being killed jules. So that's the delta H. Of this reaction. Let's go ahead and solve for our delta S. So the same exact method. But the equation is slightly different. So for our change of entropy we have it's being equal to the entropy of my reactant minus the entropy of my product. So similar to our delta H but just switched. So my delta S is going to be the C. So we have two moles of our first product, the corresponding entropy value is 109.3 units being jewels per mole times kelvin. Then my second reactant is going to let's see we have three moles of this. The correspondent entropy value is 1 30.58 jewels per mole times kelvin. No subtracted from the entropy of my products. First product, we have two moles of this, the value is 28.32 units again being jewels per mole times kelvin. Then for my second product we have six moles of this and the corresponding entropy value is 86.69 Again, units is just the same. So just leave it off or else it will be cut off from the screen here, but everything will cancel nicely in terms of the moles giving us jules per kelvin. So we solve or put the numerical values into a calculator. We can see that the change of entropy or entropy will be negative 566.68 jewels Per Kelvin. Let's go ahead and convert this jules unit into killing jewels. We can do a dark conversion for every 1000 jewels that we have. We have one kill a jewel. So again, putting this into my calculator, I get that my delta S is equal to, let's see negative 0.56668 killer jewels per kelvin. And in its correct significant figures, then my delta S is equal to negative 0. killer jewels per kelvin. And this is my change of entropy for this reaction. Now finally solving for my delta G. We're gonna go ahead and relate this with our Gibbs free energy equation. But of course it's first convert our temperature then into kelvin's temperature given to us is 25 degrees Celsius. We can convert it into kelvin's by adding it with 73.15 giving us the temperature to be 298.15 kelvin's. So if you recall the gibbs free energy equation is delta G. Equaling to the change of entropy minus the temperature. So multiplied by our change of entropy. Of course we saw for all these values, we just go ahead and plug this in SAr delta H. We said to be negative 857.4 killer jewels. Their temperature is 298.15 kelvin's and their delta S is zero point or negative 0. units being killed, jules per kelvin. So then putting this into a calculator, get that minority value is negative 0.444358 units being killed jules. And of course to have these correct amount of significant figures will go ahead and use scientific notation. So that's just negative 6.88 times 10 to the second power and units being again just killer jewels. And that's going to be my delta G for this um equation. And that's it guys. Thank you all so much for watching.

Use data in Appendix C to calculate ΔH°, ΔS°, and ΔG° at 25 °C for each of the following reactions. (c) 2 P1s2 + 10 HF1g2¡2 PF51g2 + 5 H21g2

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