Consider the following reaction between oxides of nitrogen: NO₂(g) + N₂O(g) → 3 NO(g) (b) Calculate ΔG at 800 K, assuming that ΔH° and ΔS° do not change with temperature. Under standard conditions is the reaction spontaneous at 800 K?

Video transcript

Hey everyone. So today we're being asked to estimate the change in standard delta G. For the following reaction between carbon monoxide and chlorine gas to form phosgene at 785 kelvin and to determine if the reaction is spontaneous or not at this temperature. So in order to do this, since we're only being asked to use temperature and standard change in entropy and standard change in entropy. And we're trying to find standard delta G. We can go ahead and use the formula that the standard delta G. Or the standard change in delta G of the reaction is equal to the change in standard entropy of the reaction minus temperature times the change in standard change in entropy of the reaction. So the first step here is to find out what is the change in standard entropy. So the reaction for standard and will be a formation which we need to use for this is excuse me sorry, The change in delta G or delta each other reaction is equal to the difference between the standard entropy of formation of formation of the products, products and the standard entropy of formation of the reactant. So with this in mind we can go ahead and use the standard entropy of formation for each of the reactant and products, which experimentally calculated values that we can find in tables and plug them into this equation essentially. So the standard entropy of formation for carbon monoxide gas, carbon monoxide gas Is equal to negative 110.5 killer jewels per mole for chlorine gas. However, the standard changing entropy information for chlorine gas cl two gas is actually zero. It is zero kg per mole. Because standard state elements or elements in their pure form as you would find them in nature. If they're not reacting with anything else, will always have a standard changes or standard entropy of formation of zero kg per mole. Finally, for phosgene, The standard entropy of formation of COCL two is negative 219.1 kg joules per mole. So, with this in mind, we can go ahead and calculate the delta H of the reaction using the standard dental piece of formation. So, standard delta H of the reaction is equal to the difference between the sentimental piece of formation of the products minus the reactant. So that will be one mole of fashion COCL two. I'm going to leave out the units so we have a little more space but there's a -219.1. And from this we will subtract one more of carbon monoxide, Which has a standard entropy information of -110. joules per mole. And we could add the one mole of chlorine gas. One more Oops! one More cl two Multiplied by zero. Since it has a change in standard entropy information That is equal to zero kg for more and you will end up with a final answer for the delta H of the reaction of negative 108. kg joules per mole. I'm sorry, -108.6 Killings Jules. My bad 108.6 kg jewels. So that's the first part. Now we need the change in delta S. And we can use a very similar process. So for delta S. For the change in entropy it's effectively the exact same thing. It is the difference in the entropy information. The standard entropy information of the products which will denote with P. And the standard entropy of formation or entropy of formation of the reactant. So again, these are experimentally calculated values and the standard entropy information for carbon monoxide. Carbon oxide gas is 197.7 jewels Per mole kelvin. More kelvin for chlorine gas It is 223.1 jewels per mole kelvin. And for phosgene COC. L two. The standard entropy information is 283. jewels. Per mole kelvin plugging all this back into our formula. We'll get that the delta S of the reaction is equal to The One More of Fashion C. OCL two. Multiplied by it's standard entropy information of the product. Again this is concerning phosgene C. O C. 02 which is 283.5. I'll leave out the units for now. And from this we will subtract the some of the change in entropy ease or formation of the reactant. So we have one mole carbon monoxide Which has a standard entropy information of 197. jules from all kelvin. And adding to that we will have one mole of cl two gas Which has a standard entropy of formation of 223. jewels from Mole Kelvin. So this will give us a final answer for the Delta S of the reaction As negative 137.3 jewels per kelvin. However, we need this to be in kill jules because our delta H is in joules and our delta G. That we will be calculating is also in kill jules. This is a really simple conversion. Remember that for every one kg jewel we have tented third jewels. So if we have 137.3 joules per kelvin Multiplied by one. Kill a jewel for every 10 to the third jewels. While our jewels will cancel out and we'll be left with negative 0. cure jewels per kelvin. So with all of this in hand we can go ahead and plug this back into our original formula. So we stated that if delta G of the reaction is equal to that delta H of the reaction minus T. The temperature in kelvin multiplied by the change in entropy of the reaction substituting in these values we get -108.6 Killer Jewels. Our temperature as mentioned in the question above was Kelvin. And the change in entropy, the change in entropy of the reaction is negative 0. 73 killer joules per kelvin. So our calvins will cancel out. They will cancel out and we'll be left with a final answer of negative 0.8 to zero killer jewels. And since this value is negative, it is negative. That means that this is a spontaneous reaction because the change in delta free and or the change in gibbs free energy is negative. Therefore, for this reaction between carbon monoxide and chlorine gas to form phosgene, the estimated delta G of the reaction is negative 8.20 or negative 0.8 to zero killer jewels. And the reaction is spontaneous at a temperature of 785 kelvin. I hope this helps. And I look forward to seeing you all in the next one.