These reactions are important in catalytic converters in...

These reactions are important in catalytic converters in automobiles. Calculate $\Delta G^{\circ}$ for each at $298 \mathrm{~K}$. Predict the effect of increasing temperature on the magnitude of $\Delta G^{\circ}$. a. $2 \mathrm{CO}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{CO}_{2}(g)$ b. $5 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ c. $2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ d. $2 \mathrm{NH}_{3}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)$

These reactions are important in catalytic converters in automobiles. Calculate $\Delta G^{\circ}$ for each at $298 \mathrm{~K}$. Predict the effect of increasing temperature on the magnitude of $\Delta G^{\circ}$.
a. $2 \mathrm{CO}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{CO}_{2}(g)$
b. $5 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
c. $2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
d. $2 \mathrm{NH}_{3}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)$

Key Concepts

Standard Gibbs Free Energy Change (?G°)

This concept involves determining the overall free energy change for a reaction under standard conditions (typically 1 atm pressure for gases and 1 M concentration for species) by using the standard free energies of formation of the reactants and products. It is calculated using the formula ?G° = ??G_f°(products) ? ??G_f°(reactants), which allows one to assess whether a reaction is thermodynamically favorable under standard conditions.

Temperature Dependence of Gibbs Free Energy

The relationship ?G = ?H ? T?S highlights how the Gibbs free energy change is influenced not only by the enthalpy change (?H) and entropy change (?S) of the reaction but also by the absolute temperature (T). Increasing the temperature can either increase or decrease the magnitude of ?G° depending on whether the reaction is entropy-driven or enthalpy-driven, thereby affecting the reaction's spontaneity.

Reaction Spontaneity and Catalysis

A reaction is considered spontaneous when ?G is negative. Although catalysts (such as those used in catalytic converters) lower the activation energy required for a reaction and speed up the rate of reaction, they do not change the overall free energy change (?G°). Hence, understanding ?G° is essential in predicting both the spontaneity of a reaction and the impact of changing conditions such as temperature.

Transcript

00:02 Here we have to calculate the delta g values of these four reactions.

00:07 And to calculate the delta g value, we first have to calculate the delta h and delta s values for each of the reactions.

00:14 And we're going to do that using two formulas, and i've already given the delta h and s values here.

00:23 So the three formulas that we're going to be using for each of these are first the delta h, delta s, and then.

00:46 And delta g.

00:48 So we're going to be using delta h which is equal to the summation n times delta h of the products minus summation n times delta h of the reactants.

00:58 Similarly for delta s and then we're going to use delta h and delta s values in the equation delta g is equal to delta h minus d times delta s.

01:08 So let's put this up here for reference.

01:12 Let's start with equation a so starting with delta h our products for the first reaction are in 2 and co2 delt age for n2 is 0 plus 2 moles of co2 is minus 393 .5 minus our reactants are co minus 110 .5 plus n0, 91 .3.

02:11 So this gives us a final delta h value of minus 748 .6 kilojoules.

02:24 X delta s reactants into 191 .6 plus two times to 13 .8, minus two times 197 .8, minus two times 197.

02:48 Plus 2 times 2 .10 .8 and gives us a final delta s value minus 197 .8 jolt per kelvin.

03:05 And now to calculate delta g minus 748 .6 kilojoules minus 298 kelvin times minus 197 .7 .8.

03:31 Converting that into kilojoules.

03:37 You can convert that into kilojoules right here by dividing it a thousand and gives us a final delta g value of minus 689 .7 kilojoules.

03:54 This is the delta g value of the first reaction.

04:02 We're going to use the same method to calculate delta g value for the rest of the equation.

04:10 So b delta h is equal to 2 times nh 3 is minus 45 .9 plus 2 times minus 241 .8 minus our reactants are 5h2, h2 is 0, 2 times an o which is 91 .3...

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