Enroll in one of our FREE online STEM bootcamps. Join today and start acing your classes!View Bootcamps

Problem 104

A key step in the metabolism of glucose for energ…

07:57
Problem 103

The molecular scene depicts a gaseous equilibrium mixture at $460^{\circ} \mathrm{C}$ for the reaction of $\mathrm{H}_{2}(\text {blue})$ and $\mathrm{I}_{2}$ (purple) to form HI. Each molecule represents 0.010 mol and the container volume is 1.0 $\mathrm{L}$ . (a) Is $K_{\mathrm{c}}>,=,$ or $<1 ?$ (b) Is $K_{\mathrm{p}}>=,$ or $<K_{\mathrm{c}} ?(\mathrm{c})$ Calculate $\Delta G_{\mathrm{rxn}}^{\circ}$ . (d) How would the value of $\Delta G_{\mathrm{rxn}}^{\circ}$ change if the purple molecules represented $\mathrm{H}_{2}$ and the blue $\mathrm{I}_{2} ?$ Explain.

Answer
Check back soon!

Topics


Discussion

You must be signed in to discuss.

Video Transcript

Okay, so first things first, we want to write out our equation. So I have a written here. We have I two plus h two going to H I, um in any time we have in equation, we want to make sure its balance. So you see, on this side we have two eyes into a ches which mean on this side, we need to put a to hear in order to balance the equation. Perfect. So were first asked if the Casey is greater than one equal to one or less than one. So the best way to do that it's defended Casey. So to rate, to write RKC expression, we're gonna do products, overreact INTs. So we're gonna put our age, I hear. And then hi to and each to here. And remember, you always raise the concentration to the power of the coefficient. So a try is going to be raised to two. I two and h two would technically be raised toe one, but we can just leave those outside will just leave it out. Perfect. So now we have to find the concentration of I two and H two and h i from the picture you're given. So if you count we have 10 H eyes in the picture to iTunes and want h two. So we're told that each molecule is 20.1 mole and it's in a container that is one leader. So in order to find the concentration, we can say 10 molecules of h I times each molecule being 0.10 moles divided by one leader. So this is our formula for concentration. Some moles over. Leader gives us polarity. So this will give us 0.1 Moeller So h eyes 0.1 Moeller And then we're gonna do the exact same thing for two. So again we're multiplying. Er for I two, we have two molecules. So times front 01 more again divided by one leader. And that is going to give us 0.2 Moeller. Now try and posit video and do, um h by yourself. So again you're gonna do the exact same thing, and you should come out with 0.1 Moeller. Okay, so next we just have to plug these into our Casey expression. I've written another one right here, So we're gonna put our h i We have a concentration of one again. Always remember to square it. And then we'll have point 02 down here and point a one. Now you just have to solve. And then you can find the Casey. So the Casey should come out to 50 so it is greater than one. All right, let's move on to the second part of the questions were asked how KP is related to K. C. So what, We can use this equation. So KP equals Casey. Times are t raised to Delta en eso delta and represents the number of moles of product minus the number of moles of reactant in the chemical equation. So we don't need to count again. We don't need to use these numbers that were given in the picture of you just need to look at these moles. So we know that there are two moles as product, and then one mole by two is a reacting in one more h two. So we know they're two moles. Um reactant it's so this would be to minus two, which gives us zero. And if we put this as zero rt to the zero raised to zero, sorry is just gonna be one. So that leaves us with KP equaling Casey every time. And the last part of the question is asking us to find the Delta G of reaction. So again, we're gonna use this equation like we always do. Well, that was not what I wanted to dio are Delta G. A reaction equals delta G of the products minus delta G up the reactant. It's so you could go ahead and plug in. I like to make little skeleton here, so we know exactly what we're plugging in Delta g of h I minus the Delta G. If I two plus the delta g of age to and remember, we want to include the coefficients. But since we want the Delta Gee of the reactions specifically for the picture that were given, we want to use these concentrations. So what, we're gonna put this 0.1 times the Delta H or the Delta G of h I. And then we're gonna do 0.2 times the delta G of I two and 0.0, 1/10 of Delta G of H two. And so again, we're doing this because usually we would use this are chemical equation So we dio a two. We use the coefficient so we do to two and then there to for h i and then one for H two and one for I two. But since we're asked about the specific mixture that were given in the picture, we want to use those concentrations that we calculated up here. So you should be able to find these values in your appendix. But I'll put them here. So this is 1.3 killer Jewell's and then time zero killer jewels times zero killer jewels. So be answer should come out to be 0.13 village rules for the Delta G of reaction. So the last part of the question asks you if the Delta do you have reaction would change if you changed. If you pretty much swamped Thies too. So the iTunes all became h twos in the H two's all became I twos and the answer is no, because both of these have, um, Delta G standards of zero. So it doesn't matter what you change this coefficient to. It's always going to be zero, so it will not affect the Delta G of reaction