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Problem 69

Consider the sublimation of iodine at $25.0^{\cir…

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Problem 68

Calculate $\Delta G_{\mathrm{rn}}^{\circ}$ for the reaction:
$$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$
Use the following reactions and given $\Delta G_{\mathrm{ren}}^{\circ}$ values:
$$\begin{array}{l}{\mathrm{Ca}(s)+\mathrm{CO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s) \quad \Delta G_{\mathrm{rm}}^{\circ}=-734.4 \mathrm{kJ}} \\ {2 \mathrm{Ca}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CaO}(s)} \quad\quad\quad\quad\quad\quad \Delta G_{\mathrm{rxn}}^{\circ}=-1206.6 \mathrm{kJ}\end{array}$$

Answer

$\Delta G_{r x n}^{\circ}=131.1 \mathrm{kJ}$


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Video Transcript

So this problem wants us to calculate Gives for energy for this reaction. Calcium carbonate going to calcium oxide plus c o d. It's gas solid, solid. And so they want you the calculus using the guy used from this reaction table that they give you. And so you can do that just by making sure you line everything up in a counter Sochi on a tree. So you look at reaction one. You see that costume carbonates on the right side instead left. So I need to flip it. And so now you have That's 1/2 0 do you? Including the face, sons? Um, because he flipped this equation your delta g of reaction. The sign flips and out 7 34 instead of negative 70 30 totals for two. You see that calcium oxide on the right side just like this one? And so the only thing in account for is the stuff counting tree, and you see that there's two calcium cz for under left side for every calcium on one, counseling on the rats out for this one. So you need to multiply everything about 1/2 to balance it out. Plus two goes to two calcium oxide also times 1/2. And so because you're multiplying both sides but 1/2 you're Delta G also gets multiplied by 1/2. Seven comes minus one too. 6.6. And from there you just add these up. So 7 34.4 plus 1/2 times negative, 7 34.4 1/2 1 to 6.6. And that gives you approximately 1 31 Tell Jules so that's your answer.

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