0:07

The question asks us to calculate the delta G at 25 degrees Celsius based on

Â the given information and decide whether or not the reaction is spontaneous.

Â We're given the balanced chemical equation.

Â We're given delta H of formation values and entropies of substances, but

Â we're not given the enthopy or the entropy of a particular reaction.

Â So the first thing we're going to have to do is calculate both the delta

Â H of formation and the delta S of the system.

Â Remember that when we're looking at our Gibbs free energy,

Â we're using the expression delta G equals delta H minus T delta S.

Â This is the formula we'll use to be able to calculate delta G and

Â determine the spontaneity of the reaction.

Â For our delta H value, we know we want to

Â have delta H of the reaction = delta

Â H of formation of the products- delta

Â H of formation of the reactants.

Â And since all of our coefficients are 1,

Â we don't have to worry about any coefficients in our calculation.

Â So delta H of reaction = delta H of our products, which is our COCl2.

Â So minus 219.1 kilojoules per mol,

Â and there's 1 mol of that,- (-110.5

Â KJ per mole for the carbon monoxide),

Â again x 1), because there's 1 mole.

Â For the chlorine, I don't need to put anything in there.

Â That's an element in its standard state, so it has enthalpy of formation of 0.

Â Then I find delta H of reaction = -108.6 kilojoules.

Â This is an exothermic process because the delta H value is negative, so

Â we're losing energy from the system.

Â 2:12

Now I need to calculate the value of delta S of the reaction.

Â Just like I did with my delta H of reaction, I'm going to look at my delta S

Â of reaction = S of the products- S of the reactants.

Â So delta S of the reaction = S of my products,

Â which I have one product, COCl2.

Â So I have 283.5 joules per mol Kelvin.

Â 2:48

And I'm going to subtract from that the entropy of

Â both of my reactants, so the entropy of CO,

Â which is 197.7 joules per mol Kelvin- the Cl2.

Â Entropies of pure substance or elements in there standard state is not 0,

Â unlike the delta H values.

Â And so, I have 223.1 joules per mol Kelvin.

Â 3:17

And what I find is that the delta S of my reaction =

Â -137.3 joules per Kelvin.

Â Because each of these is 1 mol, my mols will have cancelled out as well.

Â So now I know my delta S of reaction.

Â I also know the delta H of my reaction.

Â Now, I can use that information with my

Â delta G = delta H- T delta S to solve for delta G.

Â And what I find is delta G =

Â -108.6 kilojoules- T.

Â I'm at 25 degrees Celsius, but I need to convert that to Kelvin.

Â So that's 298 Kelvin x delta S,

Â which is -137.3 joules per Kelvin.

Â But I also need to divide by 1,000 because I want to convert that to kilojoules, so

Â that my entropy value and my enthalpy value are both with respect to kilojoules.

Â Now I have delta G = -67.7 kilojoules.

Â And because delta G is a negative value,

Â this is a spontaneous process at 25 degrees Celsius.

Â