Calculate the change in Gibbs free energy for each of the sets of...

Question

Calculate the change in Gibbs free energy for each of the sets of 
ΔH 
rxn, ΔSrxn, and T given in Problem 42. Predict whether or not 
each reaction is spontaneous at the temperature indicated. 
(Assume that all reactants and products are in their standard 
states.)

Accepted Answer

Hey everyone today, we're given four different reactions or four different conditions for a reaction. And were asked to determine if the reactions are spontaneous or not at the given temperature. So we're using delta H. The change in entropy delta S. The change in entropy and t the temperature in kelvin to try and find the change in gibbs free energy for each set of data values. So this means that to find the gifts free energy, we need to use the formula that delta G. The reaction is equal to delta H minus T delta S. Of the reaction. Keep in mind that delta G is calculated in killing jewels. This is important because we need to do one important conversation sorry not conversation conversion before going on and proceeding with the problems so we can see that our delta H. R. Change in entropy. These are all given in killer jewels and our temperatures are all given in kelvin which is great for us. However, our entropy these are given in jewels Per mole kelvin. We need to convert this to kill a jewels from all kelvin. Lucky for us they're all the same value which means we can do it once and we have it for the rest of them as well. Let's really quickly do that To convert negative 1 jewels per mole kelvin. We need to multiply by the conversion factor because recall for everyone, kill a jewel, We have 10 to the third tools so our kill it or our jewels will cancel out and we will be left with an answer Of -0.124 killer jewels per mole kelvin. So keep this value in mind because we're going to be using this for the rest of these as well. So taking a look at the first set of conditions, if were to calculate delta G. Using this, delta G is therefore equal to Our Delta H, which is negative 87 Minour the temperature, which is 925 Kelvin Times the change in entropy which is negative 0.124 joules per mole. Oh which means our final answer. Our delta G here will be Positive 27. Kill it jules. And since it is positive, this therefore means that the first reaction is non spontaneous because remember positive delta G equals non spontaneity. Delta G equals zero means we're at equilibrium and a negative delta G means that the reaction is spontaneous. Let's go ahead and take a look at the second set. The entropy has not changed. Neither has the entropy. However, the temperature has decreased. So let's write this out Is equal to negative killer jules, Permal minus four, Kelvin times negative 0.124. Kill a jules per mole, which equals negative 30. kg jewels. So since this value is negative, we're dealing with a spontaneous reaction spontaneous. Moving on to the 3rd Dataset, nothing has changed again except for the temperature. So, plugging that in, we have Delta G Is equal to negative 87 killer jewels. Permal minus two, Kelvin, which is actually standard temperature. That is 25°C times negatives .1-4 killer jules Permal. And this gives us a value of negative 50 killer jewels. Again, this is negative, which means that option three is also spontaneous. And finally Taking a look at option four, we have our delta H. Is now positive, So we need to adjust accordingly. So this is positive, 87 kg joules per mole minus 3 50 kelvin times negative 0.124 killer jewels per mole. With a final answer of positive 130 killer jewels. Since the value is positive, that makes option four non spontaneous. So There you have it. Option one is non spontaneous, two and three are spontaneous and four is also non spontaneous. I hope this helps. And I look forward to seeing you all in the next one.

Calculate the change in Gibbs free energy for each of the sets of ΔH rxn, ΔSrxn, and T given in Problem 42. Predict whether or not each reaction is spontaneous at the temperature indicated. (Assume that all reactants and products are in their standard states.)

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