Consider the reaction:
CO2( g) + CCl4( g) ∆ 2 COCl2( g)
Calculate...

Question

Consider the reaction:
CO2( g) + CCl4( g) ∆ 2 COCl2( g)
Calculate ΔG for this reaction at 25 °C under the following
conditions:
i. PCO2 = 0.112 atm
ii. PCCl4 = 0.174 atm
iii. PCOCl2 = 0.744 atm

Accepted Answer

Hey everyone today, we're being asked to calculate the Delta G. of the reaction at 25°C with the following pressures of the reactant and products. So since we're dealing with non standard gibbs, free energy, we can go ahead and assume that we're going to be utilizing the formula that delta G. Of the reaction is equal to the standard delta G. Of the reaction plus R. T. Lawn Q. Or natural log of Q. Where R. Is the universal gas constant, T. Is temperature in kelvin and Q. Is the reaction quotient. Now, before going anywhere else, let us go ahead and calculate the reaction quotient. So the reaction quotient is simply the let's write this out here. Reaction question is simply the products or the partial pressures of the products raised to their coefficients to the co's over the reactant. It seems pretty simple. So writing that out Q. Is therefore equal to 0.650, which is the C two H 50. H. Or ethanol divided by um Sorry, and times 0.114. And we're not raising any coefficients or rather we raise into a coefficient of one because each reactant and products has a coefficient of one. There's only 1 to 1 to one molar ratios. So the reaction quotient is effectively equal to 30.8203 80. M 0.8203 atmospheres. We also need to find the standard delta G. However, and this can be calculated using standard for Gibbs, free energies of formation two G. So it will be equal to the standard delta G. Of formation of the products. I will write that with p minus the standard delta G. Of the reactant. So for ethylene gas or C two H four of C two H four, it has a standard delta G. Affirmation of 68.4 kg joules per mole for water. Let's just put the F. There for water. It is negative 2 28.6 kg joules per mole. And for ethanol Gee C two H 50 H. The delta G information is negative 167.9 kg joules per mole. So, plugging that back into the uh reaction given up here. This is therefore equal to And let's just draw an arrow. We can leave out the we don't really have to worry about moller coefficients since it's all one mole times negative 1, 67.9 killed joules per mole. That is our product, our ethanol's delta G information. And from here, I'm just going to leave out the units for the sake of brevity but minus the one more times 68.4 kg jewels or leave that out. Right. Plus, because we have to take the some of the products and subtract that from or some of the reactions and subtracted from the products. One more Times -228.6. And this will give us a final standard delta G of the reaction of negative 7.7. Kill oh, jewels per mole. Let's scroll down a bit. So now that we have all our values, we can go ahead and plug this into the equation that we used earlier or that we wrote down earlier. So we have delta G. Is equal to the standard delta G. Of the reaction plus R. T. Lawn natural log of Q. And remember our temperature needs to be in kelvin And we have a 25 degree Celsius temperature which means our temperature. And let me just write this here. Before I forget It would be 25°C plus 2 73.15, which is equal to 298. 15 kelvin. Substituting all of these values into this equation, we have that our standard delta G is -7.7 killer jewels. Permal. R. As a gas constant which is in jewels per mole kelvin. But we want our answer to be in killer jewels. So we need to multiply this by a conversion factor. It's like so more carbon. The conversion factor is pretty easy because remember we have one kill a jewel for every 10 to the third jewels. So our jewels will cancel out times our temperature, Which is to 98.15 Kelvin multiplied by the natural log of 30.8203 A T. M three A. T. M solving this or simplifying it. We get a final answer that the delta G of the reaction, delta G of the reaction is equal to 0.798 kg jewels Permal. That is our final answer, I hope this helps, and I look forward to seeing you all in the next one.

Consider the reaction: CO2( g) + CCl4( g) ∆ 2 COCl2( g) Calculate ΔG for this reaction at 25 °C under the following conditions: i. PCO2 = 0.112 atm ii. PCCl4 = 0.174 atm iii. PCOCl2 = 0.744 atm

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