Use Hess's law and the following data CH4(g)+2 O2(g)...

Use Hess's law and the following data $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta_{\mathrm{r}} H^{\circ}=-802 \mathrm{kJ} \mathrm{mol}^{-1} \\ \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \\ \Delta_{\mathrm{r}} H^{\circ}=+247 \mathrm{kJ} \mathrm{mol}^{-1} \\ \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \\ \Delta_{\mathrm{r}} H^{\circ}=+206 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned}$$ to determine $\Delta_{\mathrm{r}} H^{\circ}$ for the following reaction, an important source of hydrogen gas $$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

Use Hess's law and the following data
$$\begin{aligned}
\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\
\Delta_{\mathrm{r}} H^{\circ}=-802 \mathrm{kJ} \mathrm{mol}^{-1} \\
\mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \\
\Delta_{\mathrm{r}} H^{\circ}=+247 \mathrm{kJ} \mathrm{mol}^{-1} \\
\mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \\
\Delta_{\mathrm{r}} H^{\circ}=+206 \mathrm{kJ} \mathrm{mol}^{-1}
\end{aligned}$$
to determine $\Delta_{\mathrm{r}} H^{\circ}$ for the following reaction, an important source of hydrogen gas
$$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

Key Concepts

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same regardless of the number of steps or the specific pathway taken, because enthalpy is a state function. This principle allows the enthalpy change of a complex reaction to be calculated by summing the enthalpy changes of simpler known reactions that add up to the overall reaction.

Enthalpy Change

Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. It is an important thermodynamic quantity that quantifies the energy difference between products and reactants, indicating whether a reaction is exothermic or endothermic.

Reaction Manipulation

Reaction manipulation involves reversing, scaling, or combining chemical equations to form a target reaction. When chemical equations are manipulated, the corresponding enthalpy changes must be adjusted accordingly, such as reversing the sign or multiplying by the appropriate factor, which is key to correctly applying Hess's Law.

Thermochemical Equations

Thermochemical equations represent chemical reactions along with their associated enthalpy changes. They are essential tools in thermochemistry because they provide the necessary quantitative details that enable the calculation of energy changes in reactions through processes like the application of Hess's Law.

Transcript

00:01 For this question, we are wanting to find delta h for this top, top balance equation right here.

00:08 And they've given us three other equations with delta h's to help us figure this out.

00:12 So what we want to do is we want to add these three different equations together to get the balanced chemical equation that they're asking us to find delta h for.

00:22 So to start out, i'm going to take equation three.

00:31 I'm going to times it by two so that way i can get the, the correct number of hydrogen and co2.

00:39 So i'm going to times it by 2.

00:41 So 2.

00:43 C .h .4 gas plus 2h2.

00:50 All right.

01:04 And i'm going to take this delta h and i'm going to times it by 2.

01:25 All right.

01:26 And then i'm i'm going to take equation one and i'm going to add these two equations together.

01:31 So equation one is methane gas plus 2o2 and delta h.

02:00 And so the delta h for this reaction is going to be negative 802 kilojoules per mole.

02:07 So i'm just going to add these two reactions together.

02:11 When we're adding a reaction together, we want to make sure that if there's something on the reactive site and if there's something on the product side, that we can cancel those two things out.

02:19 So i see that we have 2h2o right on this side, and then 2h2o on that side, so i'm just going to cross those out.

02:26 So we don't have to include those in our final balanced equation.

02:31 So i'm just going to go down and add the reactant in the product side.

02:34 So we're going to have 3 methane gas plus 202.

02:43 And on our product side, we're going to have co2 gas, plus two and then two carbon monoxide plus six hydrogen gas all right and then the delta h for this one we're just going to add the two delta hes for these reactions together so the delta h is going to be negative 390 kilojoules per mole all right so i've added reaction three and one together so now i'm going to take this new chemical equation that i have and i'm going to add it to equation two.

03:26 So i'm just going to rewrite this equation out.

03:43 All right.

03:44 And then we know the delta h for that one...

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