Tell whether reactions with the following values of ΔH and ΔS are spontaneous or nonspontaneous and whether they are exothermic or endothermic. (a) ΔH = - 48 kJ; ΔS = + 135 J>K at 400 K (b) ΔH = - 48 kJ; ΔS = - 135 J>K at 400 K (c) ΔH = + 48 kJ; ΔS = + 135 J>K at 400 K (d) ΔH = + 48 kJ; ΔS = - 135 J>K at 400 K

Video transcript

Welcome back, everyone. The following reactions occur at the same temperature of 25 Celsius. Given the following entropy change and entropy change values are the reactions spontaneous or non spontaneous. Determine if the reactions are endothermic or exothermic. Let's begin with reaction one. We have a given entropy change of negative 2 25 Jews per Kelvin and an LP change of negative 70 kilojoules. So let's begin by recalling our Gibbs free energy formula where we would calculate Gibbs free energy change for a reaction and recall that it's equal to our entropy change of our reaction. Delta H subtracted from the temperature in Kelvin times delta S our entropy change. Next recall that our unit for Gibbs free energy change should be in kilojoules. So beginning with example one, we'll do the work for that below. We have our gives free energy change which we need to find equal to our entropy change given as negative 70.0 kilojoules subtracted from our Kelvin temperature. So we would take 25 to Celsius and add 2 73.15 and we would get our Kelvin temperature equal to 2 98.15. Kelvin So plugging that in, we have minus 2 98.15 Kelvin multiplied by our entropy change for our reaction and we'll use a different color. So for reaction one, the entropy change is given as negative 225 with units of jewels per Kelvin. Now, as we stated, we need our final unit to be in kilojoules for gibbs free energy change. And since we have jewels per Kelvin in our entropy change, we're going to multiply it by a conversion factor to go from jewels in the denominator to kilojoules in the numerator where we would recall that our prefix kilo tells us that for one kilojoule, we have 10 to the third power of our base unit Jews. So canceling out jewels were left with kilojoules per Kelvin for our entropy change in which we can now cancel out our units of Kelvin in our multiplication here. And then we're going to take the product of our terms in parentheses. So in our next line, we have our gives free energy change equal to negative 70.0 kilojoules subtracted from our product which results in a value of negative 67.0838. And our units are kilojoules. So from this difference of these two terms, we have our gibbs free energy change for reaction one equal to negative 2.92 kilojoules. Notice that because our Gibbs free energy change is negative, this means that therefore reaction one is going to be spontaneous. And because our entropy change for reaction one, given as negative 70 kilojoules as negative, this value being negative tells us that our reaction is also exothermic. And so this would be our first answer, for example, one or reaction one. Now let's move on to reaction two. So for reaction two, we again follow the same formula formula, our gifts free energy change formula equal to our given entropy still given as negative 70.0 kilojoules subtracted from our temperature 2 98.15. Kelvin multiplied by our entropy change for a reaction two, given us positive 225 joules per Kelvin in which we multiply by our conversion factor to go from one kilojoule equivalent to 10 to the third power Jes. So again, canceling out Jews and canceling out Kelvin, we're left with kilojoules and we get the following result where we have negative 70 kilojoules subtracted from the results of our products here in parentheses which is positive 67.0838. And this was this difference results in our gives free energy change reaction two equal to negative 137 points. Or rather we would stop there since we just need 36 fix. So one negative 137 kilojoules. And so therefore, this magnitude being negative for our gives free energy change means that we have a spontaneous reaction again. And based on the also negative entropy change which is still negative 70 kilojoules. It's also an exothermic reaction. And so this would be our next two answers. Now let's move on to example, three. So example, three calculating gibbs free energy change, we have this time positive 70 kilojoules as our entropy change subtracted from 2 98.15 Kelvin, our temperature multiplied by our entropy change for reaction three positive 70 kilojoules, sorry about that. Um Our entropy change is not positive 70 Kjos. It's negative 225 tools per Kelvin as given in the prompt and then multiplying by our conversion factor to cancel out jewels in the denominator. We have one kilojoule 10 to third power jewels canceling out jewels and canceling out Kelvin. We're left with kilojoules again. We have now positive 70 kilojoules subtracted from the results of our product which is now negative 67.0838. And now we would just take the difference here, which is really just adding these two since we have two negatives. And so our gifts free energy for reaction three is positive 137 kilojoules. And therefore, that means that this reaction is non spontaneous. And this would be our third answer as well as given the positive entropy change. We know that the reaction is going to be now endothermic. Sorry. And so these are our next two answers. Now we just have reaction four to consider. So we'll try to fit that here. So for reaction four, we want to calculate the Gibbs free energy change equal to our entropy. For reaction four, positive 70 kilojoules subtracted from 2 98.15. Kelvin multiplied by our entropy change positive 225 Jews per Kelvin and multiplied by our conversion factor one kilojoule per 10 to third power jewels, canceling out jewels and canceling out Kelvin. We're left with killer jewels again and we have our gibbs free energy change equal to 70 kg subtracted from the results of our products positive 67.0838. And we will have a gives free energy change equal to positive 2.92 kilojoules. And so this would be, or we can say therefore, since this is a positive magnitude for gives free energy change of reaction four, this is a non spontaneous. Now, looking at our entropy change for reaction four, notice that it's positive in magnitude as just positive kilos. And so therefore, our reaction we can understand is going to be endothermic. And so this would be our last or final answer for part four of this prompt. And so all of our answers highlighted in yellow correspond to choice c in the multiple choice as the correct answer. And I'll describe the spontaneity of each reaction one through four and whether the reaction is exothermic or endothermic. So I hope that this helped and let us know if you have any questions.