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Manhattan College

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Problem 76

Calcium carbide $\left(\mathrm{CaC}_{2}\right)$ reacts with water to form acetylene
$\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)$ and $\mathrm{Ca}(\mathrm{OH})_{2} .$ From the following enthalpy of reaction data and data in Appendix $\mathrm{C},$ calculate $\Delta H_{f}^{\circ}$ for $\mathrm{CaC}_{2}(s) :$
\begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Delta H^{\circ}=-127.2 \mathrm{kJ} \end{aligned}

Answer
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## Video Transcript

we're doing Problem number seventy six and Chapter five Chemistry Center Science, right? Given this equation and the age values we have here, they want us to do you find the that each value for calcium carbide, remember? Has his logged out. The age is equal to that age. Reactions equal to Dottie's problem is dealt H of reactors acting. So that means native one twenty seven point to kill a Jules hold. Is it point to kill? Jules is equal to too negative. Two, two hundred and twenty six point seven seven. Thanks. Let's just get with the units that will be get Got more space seven minus nine. Ready nine. Eighty six point two minus two times and it's two eighty five twenty three. Three plus us are variable, which is going to be X. That one twenty seven point two is equal to minus seven. Nine. War three three The B and the negatives cancel out. So it's five. Seven one, six six. What's next? So one forty seven point two legal to negative one eighty seven. Take that. It's seven seven, seven plus Thanks. So let me, us and us forgot to distribute the negative signs about. Thanks. So then then we're going to have a situation we bring. We had both sides by one point seven point two. Then add both sides by X, too. And we're going to have a answer of sixty point five seven, minus sixty. Negative. Sixteen point five seven Killer. Jules, that's going to be detailed to H for our our calcium carbide here.