Use standard enthalpies of formation to calculate ΔH °rxn for
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Question

Use standard enthalpies of formation to calculate ΔH °rxn for 
each reaction. 
c. 3 NO2( g) + H2O(l )¡2 HNO3(aq) + NO( g)

Accepted Answer

Hi everyone for this problem, it says ammonia is the raw material used in the commercial production of nitric acid. The first step of the process is the oxidation of ammonia and were given the chemical equation Using standard entropy of formation, we need to calculate the standard entropy change for this reaction. So let's go ahead and write out what that means. And so the standard entropy change of our reaction is going to equal the sum of our products, minus the sum of our reactant. And I'll explain what that means. And so we're going to need our standard heats of formation for our each thing in our reaction in order to calculate this. So let's go ahead and look up those values because it's going to be necessary. And so the standard heat of formation for our ammonia is negative 46. killer jewels per mole. The standard heat of formation for R. N. O. Is going to equal positive 90.29 killer jules Permal. And the standard heat of formation for H20 gas is going to equal negative 20 sorry, negative 241 .8 Killer Jewels per more. And we need to pay attention to see if we have anything in its elemental state because anything in its elemental state Is going to equal zero. So for this problem we have oxygen and its elemental state. And so that heat of formation is zero. So let's continue with solving for our standard entropy change of our reaction. So we have The some of our products. And so let's take a look at our products, we have no gas and H20 gas. And so we're going to multiply how many moles we're going to take the some of our products minus the sum of our reactant. So we're going to take how many moles we have of each of our product and multiply it by its standard heat of formation. So let's start off with our H 20 gas first. So we have six moles of H 20 gas and we're going to multiply it by its standard heat of formation, which is negative 0.8 killer jewels Permal. And we're going to add because we're taking the some of our products, we have four moles of N. O gas, Multiplied by its standard heat of formation is 90 0.29 Killer jules Permal. So that's the sum of our products minus the sum of our reactant. And so here we said that oxygen is in its elemental state. So we can still write it out but we know it's going to be zero. So we have five moles of oxygen times zero kg joules per mole plus Formals of Ammonia and its standard heat of formation is negative 46.1 killer jewels per mole. Okay, I'm running out of space here and so we can go ahead and continue to solve this. We know that our five moles of oxygen is going to equal zero so we can go ahead and eliminate that now. But we wrote it out just so that we can see and so when we simplify this, we get our standard and they'll be of our reaction is going to equal. When we simplify the some of our products, we get negative 1000 89 point 64 killer jewels per mole minus -184 Point four killer jewels. So that's the sum of our products minus the sum of our reactant. And we get a final standard. And there'll be change of our reaction to be -905. killer jewels per this is our standard entropy change for this reaction. That's the end of this problem. I hope this was helpful.

Use standard enthalpies of formation to calculate ΔH °rxn for each reaction. c. 3 NO2( g) + H2O(l )¡2 HNO3(aq) + NO( g)

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