a. From a consideration of the following reactions, calculate ΔHf^∘ for methane, CH4(g) CO2(g)+2

A. From a consideration of the following reactions,...

a. From a consideration of the following reactions, calculate $\Delta H_{f}^{\circ}$ for methane, $\mathrm{CH}_{4}(g)$ $\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{HCHO}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=35 \mathrm{~kJ}$ $\mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) ; \Delta H^{\circ}=393 \mathrm{~kJ}$ $\mathrm{HCHO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=-201 \mathrm{~kJ}$ $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=-484 \mathrm{~kJ}$ b.From appropriate entropy data, calculate $\Delta S^{\circ}$ for the formation of methane from the elements. c.Using the information from above, calculate $\Delta G_{f}^{\circ}$ for methane at $25^{\circ} \mathrm{C}$.

a. From a consideration of the following reactions, calculate $\Delta H_{f}^{\circ}$ for methane, $\mathrm{CH}_{4}(g)$
$\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{HCHO}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=35 \mathrm{~kJ}$
$\mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) ; \Delta H^{\circ}=393 \mathrm{~kJ}$
$\mathrm{HCHO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=-201 \mathrm{~kJ}$
$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H^{\circ}=-484 \mathrm{~kJ}$
b.From appropriate entropy data, calculate $\Delta S^{\circ}$ for the formation of methane from the elements.
c.Using the information from above, calculate $\Delta G_{f}^{\circ}$ for methane at $25^{\circ} \mathrm{C}$.

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is crucial for properly balancing chemical equations and ensuring that the thermochemical data (such as enthalpy and entropy changes) are applied in the correct proportions when combining reactions or comparing different thermodynamic processes.

Gibbs Free Energy

Gibbs free energy (?G) combines the concepts of enthalpy, entropy, and temperature to predict the spontaneity of a reaction. It is defined by the equation ?G = ?H - T?S. A negative ?G indicates a thermodynamically favorable reaction, and calculating ?G for the formation of a compound is essential for understanding its stability under given conditions.

Entropy

Entropy (S) is a measure of the disorder or randomness in a system, and its change (?S) during a reaction reflects the change in molecular organization. Calculating the entropy change for a reaction, especially the formation of a compound from its elements, involves using standard molar entropies and the stoichiometry of the reaction to understand the dispersal of energy and matter.

Standard Enthalpy of Formation

The standard enthalpy of formation (?Hf°) is defined as the enthalpy change when one mole of a compound is formed directly from its elements in their most stable forms under standard state conditions. It is a key thermodynamic quantity used to understand the energy content of a substance and to compare the stabilities of various compounds.

Hess's Law

Hess’s law is a fundamental concept in thermochemistry stating that the total enthalpy change for a reaction is independent of the reaction pathway. This means that if a reaction can be expressed as a series of steps with known enthalpy changes, the overall enthalpy change is the sum of those stepwise changes. This principle is critical when calculating unknown enthalpies, such as the formation enthalpy of a compound.

Thermochemical Equations

Thermochemical equations are balanced chemical equations that include the enthalpy change for the reaction. They allow for the quantitative analysis of energy changes during chemical reactions. By appropriately manipulating these equations—such as reversing their direction or scaling them by a coefficient—one can derive unknown energy changes using known reaction data.

Transcript

00:01 So for this question, we are first trying to calculate the standard enthalpy formation for methane.

00:08 So for this, we're going to use hess's law, where we combine all of these equations to get our final equation.

00:16 So the equation for the formation of methane should be solid carbon plus hydrogen gas will form methane gas.

00:37 So this is the equation that we are trying to get.

00:42 So we're essentially going to add and put together all these other equations to get this final equation, and that'll help us figure out the change in enthalpy.

00:54 So when we're looking at these equations, we want equations to have the methane on the right hand side, and then the carbon and hydrogen on the left hand side.

01:09 So if we look at this first equation, the carbon dioxide plus the hydrogen gas goes to the two other products.

01:20 We will keep that the same because we want hydrogen gas to be on the left.

01:24 So we'll write that out.

01:25 I'm not going to include the phases to make it just a little bit more clear so it's not so cluttered.

01:37 And its change in enthalpy for that equation is 35 kilojoules.

01:43 So we didn't do anything to this equation, right? we didn't flip it, we didn't add anything to it.

01:47 So this change in enthalpy stays the same.

01:51 So now let's go to the next question where carbon dioxide is being broken up into carbon and oxygen.

01:56 We want carbon to be on the left hand side of the equation, right? because it's on the left hand side here.

02:04 So we're going to flip that equation.

02:07 We're going to do the reverse equation.

02:10 So we're going to have carbon plus oxygen gas going to carbon dioxide.

02:19 Now the change in enthalpy for that equation is 393 kilojoules, but because we reversed it, we changed the sign.

02:29 So instead of positive 393 kilojoules, it's going to be negative 393 kilojoules.

02:35 So then let's go to the next question, or the next equation.

02:42 And we have hydrogen gas on the left and methane on the right.

02:45 So we're going to keep that equation as is because that's where we want them.

02:54 So we have our hydrogen gas on the left, which is what we want.

02:56 And then we have our methane on the right, which is what we want.

03:01 And then because we kept that equation the same, the change in enthalpy, negative 201 kilojoules, is also going to stay the same.

03:10 And then if we look at the last equation, we have hydrogen gas on the left, oxygen on the left, and then water on the right.

03:26 However, when we are looking at these other equations, so if you look at the equations we've already written, we already have our carbon and we have hydrogen on the left and we have our methane on the right.

03:49 So we're good there.

03:51 But if you look, we have two waters on the right and nothing on the left to cancel them out.

03:57 So for this last reaction, we're going to use the reverse reaction.

04:03 So that way water's on the left, so it can cancel out with these two waters.

04:08 And then the hydrogen gas and oxygen will be on the right, so they can cancel out with these two components here.

04:15 So let me write that out.

04:17 So the last reaction, we're going to do it in the reverse.

04:20 So two waters, it's going to go to the hydrogen gas and oxygen gas.

04:29 The change in enthalpy is negative 484 kilojoules, but because we reversed the reaction, we do the opposite sign...

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