For the fission reaction #2""_(1)^(2) "H" -> ""_(1)^(3) "H" + ""_(1)^(1) "H"#, where the isotopic masses are #"2.01410 amu"#, #"3.01605 amu"#, and #"1.00782 amu"# respectively, what is the change in energy for this reaction in #"J/mol"#?
Hint: first calculate the mass defect, and then use #E = mc^2# .
Hint: first calculate the mass defect, and then use
1 Answer
#DeltaE = -3.892 xx 10^11 "J/mol"#
You should follow the hint. In this case, we should first find the initial and final masses.
#m_(""_1^2 H) = "2.01410 amu"# , for each deuterium atom on the reactants' side.
#m_(""_1^3 H) = "3.01605 amu"# , for the one tritium atom on the products' side.
#m_(""_1^1 H) = "1.00782 amu"# , for the hydrogen atom on the products' side.
The initial mass is:
#m_i = sum_k m_(k,i)#
#= 2 xx m_(""_1^2 H) = "4.02820 amu"#
And the final mass is:
#m_f = sum_l m_(k,f)#
#= m_(""_1^3 H) + m_(""_1^1 H) = "4.02387 amu"#
So, the change in mass is:
#Deltam = m_f - m_i = 4.02387 - 4.02820#
#= -"0.00433 amu"# , or#"g/mol"# .
And that is your mass defect. The change in energy associated with this mass defect is then going to use the change in mass in
#color(blue)(DeltaE) = Deltamc^2#
#= (-0.00433 xx 10^(-3) "kg/mol")(2.998 xx 10^(8) "m/s")^2#
#= color(blue)(-3.892 xx 10^11 "J/mol")#
