Given the following data Ca(s)+2 C( graphite ) ⟶CaC2(s) ΔH =-62.8 kJ Ca(s)+(1)/(2

step 1 All right, so we have our reactions here. We're going to kind of erase and go as we're figuring

Given the following data Ca(s)+2 C( graphite ) ⟶CaC2(s)...

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate $\Delta H$ for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

Given the following data
$$
\begin{aligned}
\mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\
\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\
\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\
\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\
\mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ}
\end{aligned}
$$
calculate $\Delta H$ for the reaction
$$
\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)
$$

Key Concepts

Hess's Law

Hess's Law states that the total enthalpy change of a chemical reaction is independent of the pathway between the initial and final states. This means that if a reaction can be expressed as the sum of several steps, the overall enthalpy change is the algebraic sum of the enthalpy changes for the individual steps. This principle is a direct consequence of the fact that enthalpy is a state function, depending only on the initial and final states of the system.

Enthalpy Change

Enthalpy change (?H) is a thermodynamic quantity representing the heat absorbed or released during a chemical reaction at constant pressure. It indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat) and plays a critical role in understanding energy transfer in chemical processes. Determining the overall ?H for a reaction often requires using known enthalpy changes of related reactions.

Manipulation of Thermochemical Equations

Manipulation of thermochemical equations involves reversing, scaling, or adding chemical equations to derive the enthalpy change for a target reaction. When a reaction is reversed, the sign of ?H is also reversed; when multiplied by a coefficient, ?H is multiplied by the same factor; and when reactions are added, their enthalpy changes are summed. This technique, based on Hess's Law, allows for the calculation of unknown reaction enthalpies using known data.

Transcript

00:01 All right, so we have our reactions here.

00:06 We're going to kind of erase and go as we're figuring out stuff we need, stuff we don't, and kind of working through.

00:12 So we've got these five steps, and we want to end with this desired problem.

00:18 So first thing i notice, the cas2 solid only appears once, which means it needs to match exactly.

00:28 So that means we're going to have to flip this reaction.

00:31 So we're going to flip this reaction around.

00:39 So that way it's on the correct side of being a reactant now.

00:44 So now we've got c -a -c -s - solid making calcium solid plus two carbon graphites.

00:59 And since we flipped it, we need to flip the sign of the delta h to now being positive.

01:08 So that takes care of the cacs 2.

01:12 The next thing i notice, well the waters are helpful, but because they're in multiple places, like there's water, there's water, there's water in a couple places, i'm going to come back to that one later.

01:28 But the c -a -o -h -2 only appears once.

01:32 So let's go ahead and make sure that's actually working out correctly.

01:35 So it's on the correct side and it has the correct coefficient.

01:40 So we actually don't need to change that problem.

01:42 It is good to go.

01:45 Now we can take a look at the c2h2 gas.

01:48 Oh, okay.

01:49 Well, that's also in the correct place, but it at least only appears once.

01:55 So we can just kind of flip the equation, the coefficients match.

02:00 So we'll flip this one around and flip the sign of our delta h.

02:10 So now we have two carbon dioxide gases plus h2o liquid making 5 halves o2 gas plus c2h2 gas.

02:32 And since we flipped it, the delta h sign also needs to flip.