Consider the following three reactions:
(i) Ti1s2 + 2 Cl21g2¡TiCl...

Question

Consider the following three reactions:
(i) Ti1s2 + 2 Cl21g2¡TiCl41g2
(a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

Accepted Answer

Hello everyone. So in this video we want to sell for our delta H. Delta G. Are constant. And our delta S. Value for this reaction over here. And of course we're given all the necessary values that we will be using to sell for each and this table. So first things first we can go ahead and solve for our adult h delta ages ego to the delta age of our products minus the delta age of our attractants. So taking a look at my reaction, then we have the S. N. C. L. Four that being my product and these two the essence and chlorine is going to be my reactant. And we have given each term here. So let's go ahead and start. So my delta H. Let's see delta age of products. We will have negative 471.5 minus, let's see zero plus two times zero. That would give us the value of -471. killer jewels. So the reason why we have to multiply by two is because we have two moles. All right now we have the delta H. Wrong color here. We have delta H. Equaling two negative 4 71.5 kg jewels. So that's one of our answers here for this problem. Now moving on to our delta H. Or delta S. So same exact pattern. We have the values of our products minus the the values for our reactant. So my delta S. Is that equal to, let's see here we have 365.8 minus 51.2 times two times the 2 to 3.1 value. You put this into the calculator we get the value of -131. units being jewels per kelvin. So my delta S value is equal to negative 1 31.6 jewels per kelvin. So that's again one of my answers for this problem. Now solving for our delta G value again. Same exact pardon the delta G of my products minus the delta G. Of the reactant. So delta G. N. Is equal to. Let's see here we have 432.2 and that's negative plus let's see, zero plus two times zero. Putting that into the calculator, we get the value of negative 4 32 points to kill a jules. So our delta G value for this equation is equal to negative 432.2 kg jewels. So this is my third answer for this problem. I'll go ahead and scroll down just a little bit. We're gonna go ahead and squeeze in the main part or the last part of this um problem. So we're trying to solve now for our cake constant. So the reaction that we will or the equation that we're going to use for that is that delta G is equal to the negative R. Of T. Multiplied by the natural log of our k constant. We're isolating R k here. So we do so with some mathematical manipulation we get that are constant is equal to E raised to the power of delta G over negative R. Of T. So we can go ahead and plug in all the values then because we have we have calculated already for all these values. So the delta G. We said to be equal to negative 432. R. Value for delta G that we saw for is in kilo jewels. We needed to be in jewels. So we can go ahead and multiply this by and units being drooled. Then R. R. Is a constant. That is equal to 8.314 jewels. Polar molt times kelvin and their T value. Let's see, so T. Is given to us in Celsius in the problem. So to convert this into kelvin's, we have 25 plus. Let's see here To 73.5 or 15. And that would give us The Kelvin's 2-98.15. So that's the value. Can go ahead and plug in then to 98.15 Kelvin's. So once I put this big fraction into my calculator, I will get the value of 174.4. No Putting again this value into my calculator, I'll get the key constant being 5.2769 times to the 75. So then round it to the correct number of significant figures. We get that K. Is equal to five times 10 to the 75 power. So this is going to be my last answer for this problem. So again, we have our delta H, delta, S, delta G, and r K constant.

Consider the following three reactions: (i) Ti1s2 + 2 Cl21g2¡TiCl41g2 (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

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