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Problem 73

Ohio State University

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Problem 70

Calculate $\Delta G^{\circ}$ at 298 $\mathrm{K}$ for each reaction:

$$\begin{array}{l}{\text { (a) } 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g) ; K=1.58 \times 10^{7}} \\ {\text { (b) } \mathrm{Cu}_{2} \mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cu}(s)+\mathrm{SO}_{2}(g) ; K=3.25 \times 10^{37}}\end{array}$$

Answer

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## Discussion

## Video Transcript

in parts A and B of this problem were given a chemical reaction and a value for the equilibrium. Constant. K. We want to calculate the value of the change in Gibbs Free Energy Delta G at standard conditions using the equation. Delta G equals negative RTL in of K. Since we're given a value for the equilibrium, constant K and we're told that the temperature is 298 kelvin, we can use this equation to directly sell for that change in Gibbs Free energy. At standard conditions, starting with Port A Delta G equals negative are which is a constant 8.314 Jules her mole times kelvin times t the temperature given us 298 kelvin times the natural log of the given equilibrium constant K, which is 1.58 times 10 to the power of seven. And again, that is unit list so we can take the natural log of it 1.58 times 10 to the seventh. And when we calculate that out, we should get a final answer of Delta G. At standard conditions, you be about negative 41 0.1 killer jewels normal because when we multiply, we can't sell out units of Kelvin. We're left with units of jewels per mole, and you can divide that by 1000 to get killing joules per mole For the final answer to part A in Part B were given another reaction with a different equilibrium. Constant value for K, and we use the same equation now to solve for filter G. It's dinner conditions for this reaction are is again eight point 314 Jules for more times, Kelvin temperature is 298 Kelvin and we take the natural log of K again. And this time the equilibrium constant is 3.25 times 10 to the power of 37. So 3.25 times 10 to the power of 37. We work through that math again and saw for a final answer of the changing Gibbs Free Energy at standard conditions to be about negative 214 killing joules per mole. That is our final answer for the reaction and per p