The standard enthalpy change for the reaciton of SO3(g) with H2O(...

Question

The standard enthalpy change for the reaciton of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH° = -227.8 kJ. Use the information in Problem 9.104 to calculate ΔH°f for H2SO4(aq) in kJ/mol. [For H2O(l), ΔH°f = -285.88 kJ/mol.]

Accepted Answer

hey everyone in this example, we're given the below data and we need to find our change in entropy for our given final reaction. So we need to figure out how we can manipulate our given data to end up with our final expression here to find our change in entropy. So looking at our first expression here or first equation, we have one mole of H two gas added to one half moles of gas producing one mole of water. Now our first step should be focused on getting rid of this 1/ mole in front of our water so we can go ahead and in order to get rid of that fraction as our coefficient, we can go ahead and multiply this entire equation by two and whatever we multiply our equation by we need to also multiply its associated entropy value. Also by that figure. So we would multiply our entropy of negative 2 85.8 kill jules by two. And this would give us a new expression where we would have two moles of H two gas added to two moles or sorry, one mole of our oxygen. So we can just write 02 gas producing two moles of water. And then our entropy value here, now that we've multiplied it by two is going to equal a value of negative 571.6 kg joules. And so now we want to go ahead and determine how we should manipulate our second step in our anthem data. And so comparing this second step in our equation here to our final equation, we would recognize that we should end up with two moles of our di nitrogen Penta oxide gas, whereas we only have one mole here. So we would go ahead and want to multiply this entire expression yet again by two here so that we can end up with two moles of nitrogen di nitrogen Penta oxide. So this means we would also go ahead and multiply our second entropy listed by two. And this would give us for our second expression. Two moles of di nitrogen Penta oxide gas Added to two moles of water producing formals of our nitric acid. And then our entropy value here is going to be multiplied by two, which is going to give us a value of negative 153.2 kg jewels. Now in our 3rd entropy expression, we should recognize that compared to our final expression here, We want to go ahead and have our nitrogen gas end up on the product side. Whereas right now it's on our reacting side still. And we also want to go ahead and get rid of these fractions because as you can see in our final expression here, we do not have any hydrogen gas. So we want to go ahead and recall that from our earlier rewritten in therapy expression we have two moles of hydrogen gas So we can go ahead and cancel out the hydrogen gas here by multiplying this expression by four and this would also allow us to have the appropriate amount appropriate amount of oxygen gas in our final expression. So because we're multiplying this expression by four, We're going to also multiply our anthill B x four. And we also want to recognize that in order to flip our nitrogen gas to the product side, we want to go ahead and flip the sign of our entropy. So we're going to reverse our entropy value. And so what we would have is now four moles of our nitric acid which produces two moles of our nitrogen gas Added to now six moles of our oxygen gas, Which is then added to two moles of our hydrogen gas. And as far as our entropy value here, this is going to be equal to negative for multiplied by negative 1 74.1 kg jewels. And so this is going to give us a value equal to positive 6 96.4 kg joules because we've reversed r expression. And so now our entropy sign is going to be positive instead of negative. So now we want to see how these rewritten expressions add up together in order to end up with our final expression and our final entropy value. So what we're going to do is cancel out our H two gas is here we have two moles on the react inside and two moles on the product side, we can also go ahead and cancel out our two moles of water with our two moles of water in the first expression, we also have four moles of nitric acid to get rid of. On the product side in the second expression. And then on the reactant side of our third expression. And we can also cancel out that single oxygen gas in our first expression with one of the oxygen gasses in our final expression, which is going to leave us with now five moles of oxygen gas, which is exactly what we have in our final expression. And so now we can go ahead and carry everything left that we have down. So what we're going to have is two moles of our nitrogen Penta oxide gas which produces two moles of our nitrogen gas and five moles of our oxygen gas because we've canceled everything else out. And so now all that's left to do is to add up all of our and their p values. So for our final entropy value we would take our negative 71.6 kg jewels and add that Two are negative 153.2 kg joules. And then we're going to add that to our last entropy which we found from our third expression That we manipulated as 6 96.4 kg jewels. And this is going to give us our final entropy value for our final expression equal to a value of negative 28.4 kg jewels. And so this would be our final answer here to complete this example for our given expression. So I hope that everything in this example was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

The standard enthalpy change for the reaciton of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH° = -227.8 kJ. Use the information in Problem 9.104 to calculate ΔH°f for H2SO4(aq) in kJ/mol. [For H2O(l), ΔH°f = -285.88 kJ/mol.]

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