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Problem 39

Calculate $\Delta S_{\text { urr }}$ at the indic…

Problem 38

Without doing any calculations, determine the signs of $\Delta S_{\text { sys }}$ and $\Delta S_{\text { surf }}$ for each chemical reaction. In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction is spontaneous.
\begin{equation}\begin{array}{l}{\text { a. } 2 \mathrm\ {CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm\ {CO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-566.0 \mathrm{kJ}} \\ {\text { b. } 2 \mathrm\ {NO}_{2}(g) \longrightarrow 2 \mathrm\ {NO}(g)+\mathrm{O}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=+113.1 \mathrm{kJ}} \\ {\text { c. } 2 \mathrm\ {H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm\ {H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-483.6 \mathrm{kJ}} \\ {\text { d. } \mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=+393.5 \mathrm{kJ}}\end{array}\end{equation}


PART $\mathrm{A} : \Delta S^{\circ}<0$ and spontaneous at low temperatures
PART $\mathrm{B} : \Delta S^{\circ}>0$ and spontaneous at high temperatures
PART $\mathrm{C} : \Delta S^{\circ}<0$ and spontaneous at low temperatures
PART $\mathrm{D} : \Delta S^{\circ}<0$ and nonspontaneous at all temperatures



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Video Transcript

So this questions little complex and has multiple parts to consider. So they want to know for these reactions. We want to find out what Delta s its surroundings is. Don't ask system and then we're the reaction spontaneous. So birthing in whether Delta G is positive or negative. And is that happen? That I had temperatures, low temperatures, all temperatures. So what value of tea does that happen at? So you look at Delta s of surroundings, We can figure that one out by using this equation or dealt as the surrounding equals minus two at the age of system ever T and so T is always positive if you're considering in Calvin and so all that matters is the sound adult age of system. So when Dr H is negative, don't s the surroundings, it's positive. And when does age is positive? Don't s subsist of surroundings is negative. And so luckily, they've given us adults age of each reaction in this in the problem itself, so we can figure that out for easy. Don't s of system will have to figure out by looking at the different components of this problem and seeing if it becomes more disorder to lesser disordered as it goes from reckons to products. And with that value and the H, we can figure out whether does you'll be positive or negative and what temperature range it will happen. So we figure that out by looking at Delta G equals that the H minus t delta? Yes. So the pain on the values of Delta s in the H the how large or small T is can affect whether it is processed this spontaneous or non spontaneous. So let's look at the 1st 1 for a dealt age of the reaction is an active member. So don't s of surroundings. It's gonna be positive. So let's figure out what don't s of the system is. Well, you're going from three moles of gas, the two moles of the gas molecule. So you're becoming less disordered to adult the ass of the system. There's gonna be negative because you're losing your decrease, Kantra. Peter's less disorder. So what does that say about Delta G? So we know. Let's line it up of our equation. Delta H Manistee don't s so in Ah h is a negative. And Delta s of a system is also negative. So negative times negative is a positive. So it's really a negative number, plus some number, Um, based on t and up the S. But mainly concerned on how large tea is. So 20 is very lit large. It overcomes the delta H value and it is positive which we don't want for a spontaneous reaction. So for this reaction to be be spontaneous, this value needs to be very small, so the negative number is larger. And so at low temperatures, Delta G equals negative and spontaneous. So as 1st 1 second when we do the same way, we were given our delta h of reaction. And that's positive. So don't s of surroundings will be negative. Don't the s of your system. We're going from two moles of gas. Three miles of gas to the entry is increasing, is growing more disordered. So this is positive. And so this is the opposite situation. We had last time still sign up with our equation again, don't they? H money's t dot the i s so that the H is a positive number and this is a positive number. So you have a positive number minus some value based on T and said If tea is very large, this number will outweigh this one. And don't you be negative So at large t large t about the G is negative and spontaneous. So for a question c same thing Delta s of surroundings and because of the h is negative that as a positive don't s of surroundings is positive. So if you look at the reaction, we're going from three moles of gas, the two moles of the gas. So we're becoming less disorders. Adult s at the system is negative. So we have the same situation as on problem one so adult at low temperatures, Dr Jean, you will be negative and spontaneous. And if you're can confuse about that, just go through and do the same process we did with this one with question one on this one and you'll see it's the exact same situation. So her 3rd 1 So for D Dr H, it's positive. Delta s of surroundings will be negative. Based on the equation I mentioned in the first page of fitness. So don't the sf surroundings is negative, So don't s of systems. So we're going from one mould gasto women will be gassed. So we're going from one component to two components, two separate molecules. So we're becoming more disordered. So I would say Delta s for the system is increasing entries increasing. You're becoming more disorder. So this is the same based on these values and these values. That's the same as question be which we did here. So large temperature Delta G is negative, and reaction is spontaneous. So for these problems, you always have toe Compare your delta H figure out your surroundings. Look at the overall change in disorder to figure out your del test of systems and with ease your guilt it to reaction. Don't chase a systems you can figure out at what temperature range it will occur based on this relationship here.

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