Chemistry, 30.11.2020 17:40 williamnason123

Which reactions have a positive ΔSrxn? 2A(g)+B(g)⟶4C(g) 2A(g)+3B(g)⟶4C(g) A(s)+2B(g)⟶C(g) 2A(g)+B(s)⟶3C(g)

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Answer from: lorelaistudent

Most likely:

${\rm 2\, A}\, (g) + {\rm B}\, (g) \to 4\, {\rm C}\, (g)$ (the first choice.)

$2\, {\rm A}\, (g) + {\rm B}\, (s) \to 3\, {\rm C}\, (g)$ (the fourth choice.)

Explanation:

$\Delta S_\text{rxn}$ denotes the entropy change in a reaction.

The value of $\Delta S_\text{rxn}$ is greater than zero if the reaction corresponds to an increase in entropy. In other words, entropy should be higher for the products of this reaction than for its reactants.

The exact entropy of each species depends on both the environment (temperature, pressure, etc.) and on the structure of the species. However, there are some general trends.

Consider one mole of some gaseous molecules and one mole of some solid particles. If that gas and that solid are under the same temperature and pressure, which one would have a higher entropy? Generally, the gas would have a higher entropy.

Also, the entropy of a gas is approximately proportional to the number of particles in that gas.

Therefore, it is likely that $\Delta S_\text{rxn}$ would be positive for a reaction if there are more gas particles among the products than among the reactants.

Note that (s) is the state symbol for solids, whereas (g) is the state symbol for gases.

In this question:

First reaction: ${\rm 2\, A}\, (g) + {\rm B}\, (g) \to 4\, {\rm C}\, (g)$ .

In each unit of this reaction, there are:

two gas particles among the reactants.four gas particles among the products. The number of gas particles that this reaction produces is larger than the number of gas particles that it consumes. Therefore, it is likely that $\Delta S_\text{rxn} 0$ for this reaction.

Second reaction: ${\rm 2\, A}\, (g) + 3\, {\rm B}\, (g) \to 4\, {\rm C}\, (g)$ .

In each reaction, there are:

five gas particles among the reactants.four gas particles among the products. The number of gas particles that this reaction produces is smaller than the number of gas particles that it consumes. Therefore, it is likely that $\Delta S_\text{rxn} < 0$ for this reaction.

Third reaction: ${\rm A}\, (s) + 2\, {\rm B}\, (g) \to {\rm C}\, (g)$ .

In each reaction, there are:

two gas particles among the reactants.one gas particle among the products. The number of gas particles that this reaction produces is smaller than the number of gas particles that it consumes. Therefore, it is likely that $\Delta S_\text{rxn} < 0$ for this reaction.

Fourth reaction: ${\rm 2\, A}\, (g) + {\rm B}\, (s) \to 3\, {\rm C}\, (g)$ .

In each unit of this reaction, there are:

two gas particles among the reactants (note that the state symbol (s)

suggests that ${\rm B}\, (s)$ is not a gas.)three gas particles among the products. The number of gas particles that this reaction produces is larger than the number of gas particles that it consumes. Therefore, it is likely that $\Delta S_\text{rxn} 0$ for this reaction.

Answer from: Quest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

Answer from: Quest

for points guys

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