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Problem 67

Calculate $K$ at 298 $\mathrm{K}$ for each reacti…

05:12
Problem 66

What is the difference between $\Delta G^{\circ}$ and $\Delta G ?$ Under what circumstances does $\Delta G=\Delta G^{\circ} ?$

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Video Transcript

and this problem, we need to explain the differences between the Change and Gibbs Free Energy Delta G at standard conditions in the changing Gibbs Free Energy Delta G at any set of conditions. The main difference between these two in terms of the symbols for these variables is that this zero corresponds to standard conditions. And this standard conditions mean that we have one Moeller of concentration units for all these species involved in the reaction. And all of the species have a partial pressure of one atmosphere. The equation that we used to find Delta G s standard conditions is negative are which is a constant times a temperature times the Ellen of K, where K is equal to the equilibrium constant of this reaction. We confined K by taking the concentration of all the gashes, iniquitous species of the products of the reaction and multiplying them and then dividing by the product of the concentrations of all of the reactant. Since of that chemical reaction. And we could also define this this K value in terms of of partial pressure units. So the personal pressure of each product divided by the partial pressure of each reactant and when When the system is that equilibrium, we know that that means that Delta G is equal to zero. So if we set Delta G equal to zero to represent equilibrium conditions at a given temperature value, then we can solve four K which would correspond to the equilibrium, concentration, the equilibrium, constant value, either k c. If we were using the concentration equation or KP, we're using the partial pressure reform of the equation for the equation to find Delta G at any standard conditions. We used the Delta G at standard conditions as a component of this equation and we add the product of Argenti multiplied by the Ellen of Q and CUE is similar to K, and this is called the the reaction quotient, and we use the same formulas to solve for Q. However, these convey any values for concentrations or pressures of the system. So those were the main differences between Delta G at standard conditions and Delta G at any given conditions. Now we want to explain under what circumstances we can have Delta G equal, Delta G at standard conditions. We look at this equation on the bottom. We see that we can have that relationship come to fruition. If this term RTL in of Q goes to zero are is a constant and T is absolute temperature and killed in which will always be positive. So the only way we can get RTL in of Q two B zero is if Ellen of Q is zero. We can only have that be the case if Q is equal to one. And so if we have Q equal toe one and Ellen of Q is zero, and this term goes to zero and Delta G equals Delta G. It's standard conditions the only way that we can have Q equal one. It's if we have one Moeller of, of all the species in the chemical reaction in one atmosphere of partial pressure for each species, so that either K c, where KP can come out to a value of one. Because in that case, the reaction quotient Q would be equal to one since we would divide the numerator, which would come out to one divided by the denominator, should also be one, so only at standard conditions again of one Moeller and one atmosphere of all the species in the reaction can Delta G equal Delta G at standard conditions