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Problem 95

Consider the following reaction: $$2 \mathrm{NOB…


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Problem 94

(a) Write a balanced equation for the gaseous reaction between $\mathrm{N}_{2} \mathrm{O}_{5}$ and $\mathrm{F}_{2}$ to form $\mathrm{NF}_{3}$ and $\mathrm{O}_{2} .(\mathrm{b})$ Determine $\Delta G_{\mathrm{rxn}}^{\circ}$ (c) Find $\Delta G_{\mathrm{rxn}}$ at 298 $\mathrm{K}$ if $P_{\mathrm{N}_{2} \mathrm{O}_{5}}=P_{\mathrm{F}_{2}}=0.20 \mathrm{atm}, P_{\mathrm{NF}_{3}}=$
$0.25 \mathrm{atm},$ and $P_{\mathrm{O}_{2}}=0.50 \mathrm{atm} .$

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Video Transcript

in part a of this problem. We're told that n 205 and f to react to form an F three and 02 in a gas phase reaction. Since we know that these air gases all of these substances will be in that state, we can set up the reaction in this way based on the information that we're given in the problem statement and a port A we just have to balance this reaction. So if we look, we can see that we need two moles of of n 205 in order to balance four moles of NF three to get or nitrogen is on each side of this reaction. After we do that, we have 10 moles of of oxygen atoms on the left side from n 205 So we need a coefficient of five here bounce out of those 10 oxygen atoms and we also see that on the product side we have a total of 12 boring Adams from four moles of NF three and so we need a striking metric coefficient of six for F to in order to balance those out. So that is the answer to part A as we have successfully balanced this gas fees reaction part B. We want to determine the changing Gibbs free energy of this reaction. This is the equation that we use in order to find that where we first calculate the the change in Gibbs free energy of the products and then subtract the total change and Gibbs free energy of the reactions so we can break up the terms for the products in reactant and starting with the products we examine the equation of Regis balance. We see that on the product side, we have enough three and 02 Remember that just like for changes in entropy, Delta H Delta G changing Gibbs free energy is similar and that the change in energy resulting in the formation of naturally occurring elements in their natural states is zero. And that is a case for die atomic gases like F two and 02 So we do not account for the species in our equation. And so the only product of interest for us is the formals of NF three. We have formals of NF three and we can look up in the appendix. What the change in Gibbs free energy information at standard conditions is for that chemical species. And when we multiply those values together cancel off units of moles were left with total Gibbs free energy change units of killer jewels and then for the reactant again, Since F two is naturally occurring, we ignore it and only examine the two moles of n 205 que moles of n 205 multiplied by its respective value for the change in Gibbs free energy of formation at standard conditions Is this the total Gibbs free energy change of the reactant? And then we take he totally gives free energy change of the products minus the total Gibbs free energy change of the reactant. And when we do that, we get a final answer for Delta G of reaction to be negative 500 69 illegals and that is our final answer for part B. And again, this is delta G of reaction at standard conditions noted by the symbol in part C. We want Teoh use that value that we just calculated in order to find changing Gibbs free energy at new reaction conditions. So this is the equation that we use in order to do that and we're told the temperature and we know our is a constant. We just sell for the change in Gibbs free energy of the reaction at standard conditions. So all we need to find is cue the reaction quotient and we can write out que in terms of the given partial pressures in part C by doing the partial pressure of the products raised to the power of their streaky metric coefficients, and then multiply that that value for each one of the products for all the products of this reaction, and then divide that same quantity for the reactant. So when we do that, our expression for Q should look like this. Where we have the partial pressure of NF three raised to the power of four. I'm supposed partial pressure of 02 raised to the power of five. Since those air the products of this reaction with their proper squeaky metric coefficients. Then we divide that by partial pressure of and 205 square times of partial pressure of F two to the power of six, which is what we can see on the denominator of our expression For Q were given values where the partial pressures of each one of these species and so we can plug in what we know into this expression to solve for Q. So 0.25 atmospheres for and if three raised to the fourth power multiplied by 0.50 atmospheres for 02 raised to the power of five. Then we divide that by 0.20 atmospheres were into 05 raised to its tricky metric coefficient of to finally 0.20 atmospheres, or F two race to its documentary coefficient of six. When we calculate out that ratio, we should get about 47 0.68 and that is going to be unit lis. Now we have a value for Q to plug into this equation to solve four Delta G of reaction under these partial pressure conditions. So Delta G of reaction equals Delta G of reaction at standard conditions, which is what we solved for in part B and again that was negative 500 69 killing jewels for more. Then we add our tln of Q. So are is a constant of 8.314 Jules, her mole times, Kelvin and we are going to want to divide that by 1000 so that we can get it in units of killing jewels. So times 10 you the negative third killer jewels for mole times Kelvin times the temperature of 298 kelvin times the natural log of that reaction quotient we just found to be 47 0.68 And again when we cancel off units of Kelvin, this is Temperature and Kelvin. We're left with units of killing joules per mole for Delta G of reaction. When we plugged at all in, we should get a final answer or delta G of reaction to come out to a value of negative five 0.60 times 10 squared Hillary jewels Ramon, and that's should be our final answer for part C.

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