How much energy (in kJ/mol) is released in the fusion reaction of 1H and 2H atoms?

Question

How much energy (in kJ/mol) is released in the fusion reaction of 1H and 2H atoms? <CHEMICAL EQUATION>

Accepted Answer

Welcome back everyone in this example, we're told that the atomic mass values of trillium and helium five and a neutron are 3.1605 A. M. U. S. Five point oh 12057 A. M. U. S. And one point oh oh 8665 AM us assuming that the following nuclear reaction occurs under a controlled fusion process. What is the energy released per mole of the reaction? We're given the mass of an electron as 5.845799 times 10 to the negative fourth power am us. And we're given our nuclear reaction below. So our first step is to calculate our nuclear mass where we need to calculate our nuclear mass of each of our re agents specifically for trillium. And for our helium five recall that to calculate nuclear mass, we take our atomic mass subtracted from our mass of our electrons. And so we want to recall that for neutral atoms which is what we have here for neutral atoms only. Our number of protons equal our number of electrons. And so recall that we get our number of protons from the atomic number. Which is this bottom left hand subscript number for each of our symbols here represented by the symbol Z. As our atomic number, which again tells us our number of protons. So right now let's calculate the nuclear mass. First of our treaty. Um so that's helium or sorry, a hydrogen atom with a mass number of three and the atomic number one. We would see that it corresponds to one electron since we have the atomic number one. So calculating its nuclear mass, we would take its atomic mass given in the prompt as 3.01605 am use. And we're subtracting it from its nuclear mass which we calculate by taking its one electron where we multiply by the Atomic mass of our electron given in the prompt also as 5.485799 times 10 to the negative 4th power am use. And so taking the product and the difference here we're going to have a result equal to 3.1550 A. M. Use. Now taking our nuclear mass for our second particle which is our helium five. We also are given its atomic number of two. So we would say Z is equal to two meaning that our helium five has two electrons. Since this is a neutral atom of helium. So calculating its nuclear mass we're going to take I'm sorry. In our first calculation this should have been a colon here. So for our nuclear mass of helium five we're going to take its atomic mass given the prompt as five 01206 a. m. use Actually that's 5.012057 a. m. use This is subtracted from its nuclear mass where we take its two electrons multiplied by the mass of an electron given in the prompt as 5.485799 times 10 to the negative fourth power am use And what we would get from the product. And this difference is a result equal to 5.01 096 am use And sorry just so this is aligned. That's 5.01096. And the decimal is down here. Now that we have the nuclear mass for each of these particles. We want to calculate mass defect. So recall that to calculate mass defect we would take our nuclear mass of our products minus the nuclear mass. Will use a shorthand here of our reactant recall that mass defect is represented by the symbol delta M. So we would say this is equal to beginning with our products going back to our prompt, our products are are helium five and our neutron particle. So we would begin with the nuclear mass of our helium which we determined above to be 5.1096 AM use added to our second mass nuclear massive. Our product for our neutron particle which we should recall is the value 1.867 AM use this is now and sorry, this is the end of our parentheses. Now we're subtracting this from the nuclear mass of our reactant which we will write out in pinks while underline it here. So for the nuclear mass of our reactant and we'll just do it below. So we have enough room. We have going back to our equation, Two moles of our trillion particles. We have one mole of treaty in particle here. In the second mole of our treaty in particle here since we have two trillion particles reacting. So we would have two multiplied by our nuclear massive treaty in which we determined above to be 3.0155 a. m. use. And so what we would get from the product of the following expression here rather is our mass effect value delta M equal to our sum of the mass nuclear masses of our products which should be six point oh 1963 A. M. Use. Subtracted from two times our Nuclear mass of our reactant which is going to yield the results of 6.031. And now we would get from this difference our mass defect equal to negative 0.1137 A. M. Use. Now moving forward to the next step in our solution, we want to calculate the energy released per mole of our reaction. So we would have delta E. For the energy change released in the reaction. And that would be Our mass defect value which we determined is negative 0.01137 atomic mass units. And this is per atomic reaction. This is then multiplied by our conversion factor. Where we're going to go from six point oh 22 times 10 to the 23rd power atomic mass units as avocados number equivalent to one g. So this is being multiplied here not to cancel out that atomic reaction unit. We're going to multiply by avocados number again as a conversion factor. Where we understand we have six point oh 22 times 10 to the 23rd power atomic reactions units. This is atomic reaction here. This is per mole of our reaction. Continuing on our conversion, we have the conversion to get rid of that Graham Unit where we understand 1000 g are equivalent to one kg and then multiplying by our next conversion factor. Or sorry, not a conversion factor but our speed of light squared because we want to recall that in this formula for our energy change. That is equal to our mass defect times the speed of light squared. So Delta MC two. So here we have our speed of light squared. Which we should recall is equivalent to the value of 2. 792458 times 10. And let's make some room here. So times 10 to the eighth Power with units of meters per second. And this is going to be squared. So we'll place another set of parentheses here. Since this is being multiplied by our previous conversion factors to yield our final results. So let's focus on canceling out units. We can get rid of our atomic reaction units since we used it with the avocados number here, we can get rid of am use we can get rid of grams. And what we're left with is kilograms squared times meter squared times seconds. I'm sorry, divided by seconds squared. And we want to recall that one jewel is equivalent to a kilogram times a meter squared, divided by seconds squared. So all of these units here are squared equivalent to one Juul, meaning that for our final answer, we're going to get a result equal to and sorry. Just to make a correction here, we need to make sure our units for grams are aligned appropriately. So grams should have actually been in the denominator. So we would have 10 to the third power grams in the denominator. And this is equal to one kg. So now we're able to appropriately cancel out grams with grams here, leaving us with kilogram squared meters squared and second squared which we know are equal to a jewel. So now in our calculators, carefully typing this all out, we would get a result of negative one point oh 22 Times 10 to the 12th power with units of jewels. And we would understand that this is per mole of our reaction. So this is going to be our final answer as the energy released per in jewels, per mole of the reaction, which is our nuclear reaction undergoing a fusion process. So what's boxed in here in yellow is our final answer. I hope everything I explained was clear. If you have any questions, leave them down below, and I will see everyone in the next practice video.

How much energy (in kJ/mol) is released in the fusion reaction of 1H and 2H atoms?

Verified Solution

Key Concepts

Nuclear Fusion

Energy Release in Reactions

Binding Energy