Use standard enthalpies of formation to calculate ΔH °rxn for
eac...

Question

Use standard enthalpies of formation to calculate ΔH °rxn for
each reaction.
a. 2 H2S( g) + 3 O2( g)¡2 H2O(l ) + 2 SO2( g)

Accepted Answer

Hi everyone for this problem we're told gaseous hydrochloric acid is oxidized by oxygen to form water vapor and chlorine gas, identify the standard entropy change of the reaction. Okay, so we're looking for standard entropy change. And in order for us to calculate the standard entropy change, we're going to first need to write out our reaction and so our reaction is going to look like this. Okay, so now that we have our reaction in order for us to calculate the standard entropy change, we're going to need the standard heats of formation for everything that's in our reaction. And so these are values that we're going to look up and each thing is going to have its own value. So the standard heat of formation of hydrochloric acid Is negative 92.30 killer jewels per mole four 02 gas. It's going to be zero kg jewels per mole and it's zero because elements in their standard states equal zero. Okay, next we have H 20 gas and this value is negative 241.8 kg joules per mole. And for chlorine gas Our value is zero because it's in its standard state. Okay, so now that we have those values in order to calculate the standard entropy change for our reaction, it's going to equal the sum of our products minus The sum of our reactant. So we're going to take these values and multiply them by the number of moles we have for each and that will give us our final standard entropy change for the reaction. So let's go ahead and start off with our products. Okay, so for our products we have two products, we have water and chlorine gas and we have two moles of each. So four our first one. Let's start off with this one. So we're going to take our two moles of water and multiply it by its standard heat of formation which is negative 241.8 kg joules per mole. And we're going to add our second reactant which is chlorine gas and this is zero. Okay, so we can just go ahead and write in zero kg joules per mole. So that's the sum of our products. And we're going to subtract the sum of our reactant. So let's take a look at our reactant. Our first reactant is hydrochloric acid and we have four moles of that. So we'll take four moles of hydrochloric acid and multiply it by its standard heat of formation, negative 92.30 kila jewels per mole plus our second reactant is gas. So this is going to be plus zero kila jewels per mole. Okay, let's move this over so we can see in full. Okay, so this is zero killer jewels Permal. So this is the sum of our products minus the sum of our reactant. Now, when we calculate this we'll get a final value That the standard entropy change for our reaction is going to equal negative 114 0. kila jules Permal. And this is our final answer. That's the end of this problem. I hope this was helpful.

Use standard enthalpies of formation to calculate ΔH °rxn for each reaction. a. 2 H2S( g) + 3 O2( g)¡2 H2O(l ) + 2 SO2( g)

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