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Problem 80

The formation constant for the reaction
$$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$
$$K_{\mathrm{f}}=5.6 \times 10^{8} \text { at } 25^{\circ} \mathrm{C}$$
(a) What is $\Delta G^{\circ}$ at this temperature?
(b) If standard-state concentrations of the reactants and products are mixed, in which direction does the reaction proceed?
(c) Determine $\Delta G$ when $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]=0.010 M,\left[\mathrm{Ni}^{2+}\right]=$ $0.0010 M,$ and $\left[\mathrm{NH}_{3}\right]=0.0050 M .$ In which direction will the reaction proceed to achieve equilibrium?

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## Discussion

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## Video Transcript

were given a reversible chemical reaction and we're told the formation constant of this reaction. And we went to answer three different questions regarding this information report. A. We want to find the changing Gibbs Free Energy Delta G at standard conditions by doing negative rt Ellen of that given equilibrium constant so we can plug in everything that we were given in this problem software that change in Gibbs free energy. It's standard conditions because we know that our is a constant of eight point 314 joules per mole Kelvin. And the temperature we're told is 298 Kelvin and we take the natural log of that given formation Constant, which is 5.6 times 10 to the power of eight. So 5.6 times 10 to the eighth power. And now units of Kelvin cancel out. So we're left with units of jewels per mole. When we do the math, we find a delta G of formation value where this reaction to be about negative 49 0.9 eligibles per mole. All you have to do is divide the answer by 1000 to convert into killer JAL's thermal. For Delta chief. So that's the answer for part A. In a party. We want to know which way this reaction will shift in order to achieve equilibrium. If all of the species are supplied at their standard molar concentrations, if if they are all at their standard concentration values, it means that they're all each one of these species would be one Moeller, one mole, her leader. And when that's the case, the reaction quotient Q. Be equal to the concentration of that product, divided by concentration of an I two plus as one of the reactors multiplied by the concentration of and H three, the other reactant to the power of six. But if all of these were one than this, would come out to a value of one. And when Q is one from this equation to find Delta G at any conditions, Ellen of Q is zero, so that term will go away, meaning that Delta G is equal to Delta G at standard conditions. So what Part B is really asking us is at standard conditions, meaning the conditions president and port A. Since we saw it, four Delta G. It's standard conditions, given this value for the formation constant. Well, the reaction proceed to the right or the left to achieve equilibrium. Do you remember that for the equilibrium constants? If K is less than one and the reaction will tend towards the reactant or left side of the chemical reaction if K equals one, there is no shift in that reaction because it is at equilibrium. And if K is greater than one, then the reaction will proceed towards the products or the right side. In order to achieve equilibrium, you can also see that it's standard conditions in part A. We calculated a Delta G value that was negative, meaning that this is a spontaneous reaction at this temperature for spontaneous reactions. They will proceed naturally without any input of end of energy towards the product side. So we can think of it that way. Or just look at how this value is much greater and one and so. Either way, we can conclude that the reaction will shift to the right in order to achieve equilibrium at standard conditions and in part C were given different Moeller concentration values or the three chemical species involved in this reaction. Since they're all acquis, we include them in that reaction quotient. Q. We do the again, the concentration of the product and i NH three and six, divided by the product of the concentration of the to react INTs and in each case, those air raised to the power of their store, key metric coefficients and I and and I in each 36 Both have coefficients of one, and NH three has a coefficient of six, which is why it is raised to the power of six in this expression. So he first solve for that value of the reaction quotient. Q. Using this expression concentration of N i N h 362 plus was given as zero point 010 Moeller concentration of an I two plus was given as zero point 00 10 more. The concentration of NH three was given as zero point 0050 more. We calculate that out. We get a reaction quotient Q of 6.4 times 10 to the 14th Power, and now we plug that in to this equation, and we use this value for Delta G. It's standard conditions that we calculate in part a saw four Delta g thes more concentration values. So Delta G equals Delta G. It's standard conditions which from heart A, we know was negative. 49 0.9 killing jewels per mole in this time, we add are eight point 314 and we will divide this by 1000. Get in units of killing jewels, her mole times, Kelvin. So we can add it with the Delta G at standard conditions. And then that's multiplied by the temperature 298 Kelvin so that the units of Kelvin cancel out when those air multiplied. So again we have killing joules per mole was killing joules per mole times the Ln natural log of that value of Q. We just saw four to be 6.4 times 10 to the power of 14. When we do that math, we should get a value of the changing Gibbs free energy at these conditions to be about 34 0.6 illegals per mole and now at thes concentrations, we want to know if the reaction will shift to the right or left to achieve equilibrium. Compared to the Delta G A standard conditions which was a negative value. Delta G that we just saw four is a positive value, meaning that at these conditions this reaction is not spontaneous and will not Percy, naturally, to the product side. So the reactant side is favored. And also we see that compared to 5.6 times 10 to the power of eight for the formation constant at standard conditions, the reactant reaction quotient increased to 6.4 times 10 to the 14th power. And so, in order to achieve equilibrium, this constant wants to decrease to you this value at standard conditions in order to to decrease that que will will decrease meaning that the reaction will shift to the left as well, which is what we concluded based on the sign of delta G non spontaneity of that reaction of these conditions. So either way, imports see, the reaction will shift to the left towards the product side for the