Consider the following reaction between oxides of nitrogen: NO2(g)+N2 O(g) ⟶3 NO(g) (a) Use data

Consider the following reaction between oxides of nitrogen:...

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix $\mathrm{C}$ to predict how $\Delta \mathrm{G}^{\circ}$ for the reaction varies with increasing temperature. (b) Calculate $\Delta G^{\circ}$ at $800 \mathrm{~K}$, assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not change with temperature. Under standard conditions is the reaction spontaneous at $800 \mathrm{~K} ?$ (c) Calculate $\Delta G^{\circ}$ at $1000 \mathrm{~K}$. Is the reaction spontaneous under standard conditions at this temperature?

Consider the following reaction between oxides of nitrogen:
$$
\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)
$$
(a) Use data in Appendix $\mathrm{C}$ to predict how $\Delta \mathrm{G}^{\circ}$ for the reaction varies with increasing temperature. (b) Calculate $\Delta G^{\circ}$ at $800 \mathrm{~K}$, assuming that $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not change with temperature. Under standard conditions is the reaction spontaneous at $800 \mathrm{~K} ?$
(c) Calculate $\Delta G^{\circ}$ at $1000 \mathrm{~K}$. Is the reaction spontaneous under standard conditions at this temperature?

Key Concepts

Standard Reaction Conditions

Standard reaction conditions typically refer to a set temperature (often 298 K) and 1 bar or 1 atm pressures, among other defined parameters, that allow chemists to tabulate and compare thermodynamic quantities like ?H, ?S, and ?G. These conditions provide a baseline for predicting how changes in temperature and other variables affect reaction spontaneity.

Temperature Dependence

Temperature plays a critical role in determining the spontaneity of a reaction through its effect on the T?S term in the Gibbs free energy equation. As temperature increases or decreases, the balance between enthalpy and entropy contributions shifts, altering whether a reaction is thermodynamically favorable. Assessing how ?G varies with temperature is essential in understanding reaction behavior under different thermal conditions.

Reaction Spontaneity

Reaction spontaneity refers to the natural tendency of a reaction to proceed without the input of external energy. This is determined by the sign of the Gibbs free energy change. By analyzing ?H and ?S along with temperature, one can predict whether a reaction will occur spontaneously under a given set of conditions.

Enthalpy (?H)

Enthalpy is a measure of the total heat content of a system at constant pressure. It reflects the energy absorbed or released during a reaction. In the context of Gibbs free energy, the sign of ?H, in combination with the entropy term, influences the overall spontaneity of the reaction.

Gibbs Free Energy (?G)

Gibbs free energy is a thermodynamic potential that indicates the amount of work obtainable from a system, and it determines whether a reaction is spontaneous. A negative ?G implies a spontaneous process under constant pressure and temperature, while a positive value means non-spontaneity. The relationship that combines enthalpy and entropy changes, ?G = ?H - T?S, is fundamental for predicting how reaction spontaneity varies with temperature.

Entropy (?S)

Entropy is a measure of the disorder or randomness of a system. In chemical reactions, an increase in entropy (positive ?S) tends to promote spontaneity, especially at higher temperatures, while a decrease in entropy (negative ?S) can inhibit the reaction. The term T?S in the free energy equation quantifies the temperature-scaled contribution of entropy to the reaction’s free energy change.

Transcript

00:01 Let's take a look at a chemical reaction and see how it might be affected or its gibbs free energy might be affected by temperature.

00:09 And so if we take some no2 and some n2 and make some n2, we want to know what is the effect of temperature on the gibbs free energy of this? well, to be able to do that, if we want to know how that's going to be affected, what we need to do is calculate the delta h's and the delta s is of reaction for this and be able to do that.

00:37 So if you recall, the delta h of reaction is going to be equal to the sum of the number of moles of the enthalpy of formation for the products minus the sum of the number of moles times the heat of, formation for the reactants.

00:59 Same thing for the entropies.

01:16 So we're going to use this equation to be able to figure out the h and the s for each of these and then be able to look at that in terms of delta g.

01:26 So when we go and do that for the enthalpy of reaction first, under standard conditions.

01:37 What we have, because remember we do products minus reactants is we have three moles and each delta h is 90 .37 kilojoules per mole.

01:49 And then we have two reactants.

01:56 We have no2, which has a value of 33 .84 kilojoules per mole.

02:00 And we have we have 102, which has a value of 33 .84 kilojoules per mole.

02:06 And we have the same, we have to do the same thing for no2, which is going to have a value of 81 .6 kilojoules per mole.

02:14 And so when we do the math on that, we're going to find that the enthalpy of reaction is 155 .7 kilojoules.

02:28 When we go do the delta s, we'll go get those values, and we've got three moles of n0.

02:40 Each has an enthalpy of 110 .62.

02:45 That's our only product, so then we go subtract the reactants.

02:50 We have one mole of n02 at 240 .45, and we have one mole of n2, and we have one mole of n2o at times 220 .0.

03:11 And we go do the delta s of react we're going to find that the enthalpy is 171 .4 joules per kelvin.

03:27 And if we convert that into kilojoules per kelvin, that's .1714 kilojoules per kelvin.

03:36 All right, so what does this tell us? so now that we have the values that we need, what does this tell us about the gibbs free energy? so remember, gibbs free energy is equal to delta h minus t delta s.

04:00 Well, we have a positive delta h...

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