The major industrial use of hydrogen is in the production of ammonia by the Haber process: 3 H2(

step 1 So problem 12 asks us first for part A. It asks us to look at this equation, and it asks us for

The major industrial use of hydrogen is in the production of...

Key Concepts

Standard Enthalpy Change (?H°)

This concept involves determining the overall heat absorbed or released during a chemical reaction at standard state. It is calculated using the standard enthalpies of formation of reactants and products, reflecting the energy required to break bonds in reactants versus the energy released while forming bonds in products. A negative ?H° indicates that the reaction is exothermic, while a positive ?H° indicates an endothermic process.

Standard Entropy Change (?S°)

Standard entropy change quantifies the difference in disorder between products and reactants at standard conditions. It is determined by summing the standard molar entropies of the products and subtracting the standard molar entropies of the reactants. Positive ?S° suggests increased randomness in the system, while a negative ?S° indicates decreased randomness.

Gibbs Free Energy (?G°)

Gibbs free energy combines enthalpy and entropy changes to determine the spontaneous nature of a reaction at constant temperature and pressure. It is calculated using the equation ?G° = ?H° - T?S°. A negative ?G° indicates that a reaction is thermodynamically spontaneous under the given conditions, whereas a positive value signals non-spontaneity.

Temperature Dependence of Reaction Spontaneity

The spontaneity of a reaction can change with temperature, given that both ?H° and ?S° are considered. By analyzing the equation ?G° = ?H° - T?S°, one can determine the threshold temperature at which ?G° transitions from negative to positive. The reaction is spontaneous at temperatures below (or above, depending on the signs of ?H° and ?S°) this critical temperature, highlighting the role of temperature in driving the direction of chemical processes.

Transcript

00:01 So problem 12 asks us first for part a.

00:05 It asks us to look at this equation, and it asks us for part a to determine delta h, delta s, and delta g for this reaction.

00:16 So, and using appendix for.

00:19 So for these types of problems, we want to look at the table that lists the delta h, delta s, and delta g's for each individual atom, right? our equation that we're going to use for all of these things overall is our products, whether that be delta h, delta s, or delta g, but our products minus our reactants always, right? so the first thing we want to do is let's start by, we can start by any one of them, but let's just say let's start with delta h, right? so we want to find the delta h of this reaction.

01:09 So the first thing we want to do is take the delta h of our products.

01:14 So we only have one product and that is ammonium here, nh3.

01:20 So we want to take our products.

01:23 So this is defined our delta age of reaction.

01:31 And this is at standard conditions because that's what it asked for.

01:36 So first we look, if you look at appendix where you'll find that, find the delta age formation for nh.

01:44 So we're going to take, since there's 2nh3, we're going to take 2 times the delta age of formation of the nh3, which if you look that up in the table is 46 is negative 46 kilojoules per mole.

02:02 And remember to write down your units for everything in this problem because that's going to be really important.

02:08 So our delta age reaction is 2 times negative 46 kilojoules per mole.

02:15 So that's our products, it's really important products.

02:17 And then we're going to subtract our reactants.

02:20 So we have 3h2.

02:24 So h2, if you look up the delta -h formation of h2 gas, make sure you're looking up gas.

02:31 You'll find that it's actually zero.

02:34 And then we're going to add the delta -h formations where n2, which is also zero.

02:45 So this whole thing cancels out to zero.

02:48 So then we're just left with our products.

02:51 So our delta -age reaction is just going to be 2 times negative 46 kilojoules per mole, which if you type that into your calculator is negative 92 kilojoules per mole.

03:23 So that right there is your delta -h for the reaction.

03:28 So then we just want to do the same thing with either delta s or delta g.

03:33 So let's just go with delta g because it'll be easier.

03:40 And then you look at delta g reaction.

03:49 So if you look up the delta g formation for nh3, again, we have two.

03:56 We're starting with products.

03:59 And the delta g for that is negative 17, kilojoules per mole.

04:12 And then again, we're going to look at the products, and we're going to find that for both h2 gas and n2 gas, the delta g of formation is zero.

04:23 And this won't always be the case, what it is for this.

04:26 So this is going to be three times zero plus zero, which is just going to be zero.

04:33 So again, our delta g now of reaction is just going to be two times our negative 17 kilojoules per mole.

04:49 And that gives us negative 34 kilojoules per mole.

05:02 So that's the delta g of reaction.

05:04 Now you could either do the same thing for your delta s of reaction or you can make it easier and use the equation.

05:17 This equation right here, so in delta g at standard conditions is equal to delta h at standard conditions minus the temperature times delta s.

05:32 And if you use this equation, since we have our delta g and our delta h of reaction already, we can just solve for delta s and then use the equation and plug in for delta s.

05:43 So first, let's solve for delta s.

05:46 So you can take this over to this side and this over to this side and get t delta s not equals delta h, not minus delta g knot.

06:05 And then we just divide both sides by the t.

06:08 So we get delta s not equals delta h, minus delta h, minus.

06:17 Delta g on all over the temperature.

06:23 So now we know our delta h of reaction and our delta g of reaction.

06:27 And we want to find our delta s reaction...

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