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Problem 68

Calculate $\Delta G^{\circ}$ at 298 $\mathrm{K}$ …

03:18
Problem 67

Calculate $K$ at 298 $\mathrm{K}$ for each reaction:
$$\begin{array}{l}{\text { (a) } \mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)} \\ {\text { (b) } \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(l)}\end{array}$$

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Video Transcript

Okay, So to solve these problems, we're going to use the equation. Delta G equals rt lnk and we're trying to find we're trying to find K. So we're going to rearrange the equation. As you can see, I divided by a negative rt to get Alan Kay on one side and then to remove the natural log, which is PLN. We're going to use E so we hav e race to the Delta G divided by r t equals K. And remember that are is a constant in its 8.314 joules per moral mole. Kelvin. So, um, we know we're were also given the temperature and, you know, our So we just have to find Delta G. In order to find K so defined Delta G. We can use the equation we have here. So am gco three going to see a three to minus plus m g two and we can find the Delta G. The reaction for this using standard Delta G values. So I wrote out the formula. What? We're going to use what we're going to use to plug in the numbers we have, as always, products minus reactant. So we got to the Delta Chief Seo to minus plus adultery of M G two plus minus the Delta G of M. J. C. 03 again, You could look up these values their standards so you could look them up in a table. But I put them all right here. So we're just gonna plug in truck through this. So we're gonna substitute this value here. This thou you here in this? How you here and that is going to come out to be 43 0.89 And that's killing joules per mole. But we want to get Jules Permal so we can put so we can substitute into our equation up here. This equation since this is in Jules Kroll. So we're gonna multiply this by 1000 which gives us 430. Excuse me. 40438900 Sorry. That's a little messy. All right, so now we're going to use our Delta G and plug it back into our equation. We started away so again, 438 900 And again, I must said it's 43890 Sorry. I'm all over the place today. Will just erase that zero perfect. And we plug and chug through this. I recommend using your calculator, Teoh, Find E to the power of and we should get K equals two point. Oh, times 10 to the negative eight. Okay, so we're gonna do the exact same thing, use the exact same equation. I'll try to write it up here again. So we're gonna use our Delta G over negative rt. And this is all going to be raised to E, which will equal K, which is what we're solving for. So again, we're gonna find the Delta G of reaction, and we can use this with standard Delta G's, so we're going to have products minus reactant. So we have the Delta G of H 20 to the Delta G of H two plus the Delta G of 02 And again you can find these in a table. But I provided them right here. So we're gonna plug and chug. So these values, they're going to go here, and this value will go here. So it's gonna come out to be 120. Might four kill a Jules Permal, But again, we wanted in jewels per mole. So it's gonna be 12 0400 Let me make sure that's right. Since I wrote it along last one. Yet there we go. And then we're gonna take this summer, employ into our equation. Sorry. It's a little messy. All right? And so we're going to get once we plug and chug through that K equals 1.27 times. 10 to the 21. There we go.