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The Balance Between Enthalpy and Entropy

Pearson
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When we talk about chemical processes, we often talk about whether or not they're spontaneous. In looking at that, we're looking at the free energy, or delta G of a reaction. For a spontaneous reaction, delta G is negative. Delta G is related to the enthalpy and the entropy by this equation. Delta G is equal to delta H minus T delta S. So for a spontaneous reaction, delta G is a negative term. If we want a spontaneous reaction, then we need a combination of delta H and delta S to give us a negative term for delta G. So in order to have this, we can have a negative delta H or something that is exothermic and a positive delta S, something that has a greater entropy. So let's look at a reaction and determine how the entropy and enthalpy play a role. So in this calorimeter here, I have water and I'm going to add sodium nitrate and measure the temperature change. So I'm adding the solid, and I'm going to stir it to dissolve the solid. So you notice the temperature went down. So can you predict what the sign of delta H and delta S would be for this reaction? So since the temperature went down, the delta H for this reaction is positive, or it's endothermic. So since we know that delta H is positive, and we know that the reaction occurred, it was spontaneous. Then delta G is negative. So we can figure out what are delta S term is. So this, delta S or the entropy term, has to be positive in order to make this negative delta G for the spontaneous reaction. So in this reaction, our delta H is positive, and our delta S is also positive. This positive delta S is indicating that we have an increase in entropy or an increase in disorder, and this makes sense because we're dissolving a solid. We're going from a solid-ordered structure to ions in solution, a more disordered solution, thus an increase in our entropy. So let's try this again for dissolving another solid. Here I have sodium acetate. Again, I'm going to dissolve this in water and monitor the temperature. [It's rising] So as you'll notice, the temperature is increasing. So what do you predict the delta H and delta S for this reaction will be? Since the temperature is rising, the delta H for this reaction is negative. It's exothermic. Since we know that the temperature has increased, our delta H for this reaction is negative, or it's exothermic, and again, we know that the reaction occurred so it's spontaneous, so we have a negative delta G. If we have a negative delta G and a negative delta H, delta S in this case we can infer as also positive, because as we saw in the last reaction, we are dissolving a solid. We're going from a structured state to ions in solution. So we're increasing our entropy. So in this reaction, we have an exothermic or negative delta H and a positive delta S to give us our negative delta G. So how can we explain that in this reaction we have a negative delta H, but in our first reaction, we had a positive delta H? In both reactions, we're dissolving a solid into ions in solution. Well, in the first case, it makes sense that we are breaking more bonds than were forming. When we have a situation where we break more bonds than we form, we have an endothermic reaction, and delta H is positive. But in the second reaction, there's something else going on. So in this second reaction with our sodium acetate, again, to dissolve this solid, we're breaking this up into ions. So we are breaking bonds, but this ion right here hydrolyzes with water. It picks up a proton to form CH3COOH. So in this case, we're also forming more bonds, and so when we form more bonds than we break, we're exothermic, or our delta H is negative.
When we talk about chemical processes, we often talk about whether or not they're spontaneous. In looking at that, we're looking at the free energy, or delta G of a reaction. For a spontaneous reaction, delta G is negative. Delta G is related to the enthalpy and the entropy by this equation. Delta G is equal to delta H minus T delta S. So for a spontaneous reaction, delta G is a negative term. If we want a spontaneous reaction, then we need a combination of delta H and delta S to give us a negative term for delta G. So in order to have this, we can have a negative delta H or something that is exothermic and a positive delta S, something that has a greater entropy. So let's look at a reaction and determine how the entropy and enthalpy play a role. So in this calorimeter here, I have water and I'm going to add sodium nitrate and measure the temperature change. So I'm adding the solid, and I'm going to stir it to dissolve the solid. So you notice the temperature went down. So can you predict what the sign of delta H and delta S would be for this reaction? So since the temperature went down, the delta H for this reaction is positive, or it's endothermic. So since we know that delta H is positive, and we know that the reaction occurred, it was spontaneous. Then delta G is negative. So we can figure out what are delta S term is. So this, delta S or the entropy term, has to be positive in order to make this negative delta G for the spontaneous reaction. So in this reaction, our delta H is positive, and our delta S is also positive. This positive delta S is indicating that we have an increase in entropy or an increase in disorder, and this makes sense because we're dissolving a solid. We're going from a solid-ordered structure to ions in solution, a more disordered solution, thus an increase in our entropy. So let's try this again for dissolving another solid. Here I have sodium acetate. Again, I'm going to dissolve this in water and monitor the temperature. [It's rising] So as you'll notice, the temperature is increasing. So what do you predict the delta H and delta S for this reaction will be? Since the temperature is rising, the delta H for this reaction is negative. It's exothermic. Since we know that the temperature has increased, our delta H for this reaction is negative, or it's exothermic, and again, we know that the reaction occurred so it's spontaneous, so we have a negative delta G. If we have a negative delta G and a negative delta H, delta S in this case we can infer as also positive, because as we saw in the last reaction, we are dissolving a solid. We're going from a structured state to ions in solution. So we're increasing our entropy. So in this reaction, we have an exothermic or negative delta H and a positive delta S to give us our negative delta G. So how can we explain that in this reaction we have a negative delta H, but in our first reaction, we had a positive delta H? In both reactions, we're dissolving a solid into ions in solution. Well, in the first case, it makes sense that we are breaking more bonds than were forming. When we have a situation where we break more bonds than we form, we have an endothermic reaction, and delta H is positive. But in the second reaction, there's something else going on. So in this second reaction with our sodium acetate, again, to dissolve this solid, we're breaking this up into ions. So we are breaking bonds, but this ion right here hydrolyzes with water. It picks up a proton to form CH3COOH. So in this case, we're also forming more bonds, and so when we form more bonds than we break, we're exothermic, or our delta H is negative.