i already have the answer to part A Unlike chemical reactions, for which mass is
ID: 524907 • Letter: I
Question
i already have the answer to part A
Unlike chemical reactions, for which mass is conserved, nuclear reactions result in slight changes in mass. When mass is lost, it becomes energy according to the equation
E=mc2
where E is the energy in joules, m is the mass defect in kilograms, and c is the speed of light (c=3.00×108 m/s). The mass defect is the difference between the total mass of the products and the total mass of reactants. The following values can be used to calculate the mass defect.
Part A
The sun produces energy via fusion. One of the fusion reactions that occurs in the sun is
411H42He+201e
How much energy in joules is released by the fusion of 1.53 g of hydrogen-1?
Express your answer to three significant figures and include the appropriate units.
9.42×1011 J
Part B
Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction
235 92U+10nfission fragments+neutrons+3.20×1011 J/atom
What mass of uranium-235 is needed to produce the same amount of energy as the fusion reaction in Part A?
Express your answer to three significant figures and include the appropriate units.
Standard enthalpies of formation
The following table provides standard enthalpies of formation for methane (CH4), carbon dioxide (CO2), and water (H2O).
By definition, the enthalpy of formation of any element in its standard state is 0 kJ/mol.
Part C
Another way to produce energy is by burning natural gas, which is mostly methane (CH4). The reaction
CH4(g)+2O2(g)CO2(g)+2H2O(g)
shows the combustion of methane. Assuming complete combustion, what mass of methane is needed to produce the same amount of energy as the fusion reaction in Part A?
Express your answer to three significant figures and include the appropriate units.
Particle Mass(g/mol) 01e 0.00054858 11H 1.00782 42He 4.00260
Explanation / Answer
B)
235 92U+10n?fission fragments+neutrons+3.20×10?11 J/atom
your value = ?E = 9.42×1011 J
find aotms of U required:
n = (9.42*10^11)/(3.2*10^-11) = 2.9437*10^22 atoms of U
mol = (t2.9437*10^22)/(6.022*10^23) = 0.0488824 mol of U
mass = mol*MW = 0.0488824*235.5
mass = 11.5118 g of U 235 required
C)
mass of methane for this energy
E = 9.42×1011 J
Q= (9.42*10^11) K
Q = n*HRxn
n = Q/HRxn = ( (9.42*10^11) ) /(HRxn
HRxn = -393.5 + 2*-285.83 - (-74.8) = -890.36kJ7mol
so
n = (9.42*10^11)/(890.36*10^3)
n = 1057999.01164 mol of methane
mass = mol*MW = 1057999.01164*44
mass = 46551956.5122 g of methane
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