Themes > Science > Chemistry > Nuclear Chemistry > Nuclear Reactions > Einstein's Equation

Albert Einstein stated that mass and energy are interconvertable. For chemical reactions, they could be treated as independent. But for nuclear reactions, the interconvertability must be considered. He said that the mass of a particle (m) is not constant. It changes with the particle's velocity (v) relative to the observer. A particle's mass is related to its velocity and the velocity of the speed of light (c), by the following equation: 
m = ((m0) / sqrt (1 - (v/c)^2))
When the particle has no velocity (relative to the observer), v/c becomes zero, and therefore, m = m0. Thus m0 is said to the particle's rest mass. The objects we observe do not move or are not moving fast enough to make any significant change in its mass because c, the speed of light, is a very large number--approximately 2.998 x 108 m/s.
This equation also shows that nothing can move at or faster than the speed of light. As an object approaches the speed of light, v/c approaches the value of 1, and m subsequently becomes much larger. But when an object does attain the speed of light, the value of v/c will become 1, and therefore the denominator of the entire fraction will become zero. Having a denominator of zero is undefined, or not possible. The mass of the object would be infinite. Therefore nothing can attain the speed of light.

The Einstein Equation

The Einstein equation, 
DeltaE = Deltamc2
shows how one could be calculated from another. The total energy of the universe and all the mass calculated as an equivalent of energy is constant. This is the law of conservation of mass-energy. Since c2 is very large (and c2 even larger--8.988 x 1016), it would take an enormous amount of energy to make even one tiny insignificant amount of matter. For example, let's take the burning of methane: 
CH4 + 2O2 ==> CO2 + 2H2O     DeltaH° = -890 kJ

The 890 kJ of energy released is also accompanied by a loss of mass. Substituting in the values given into the Einstein equation gives us: (In this equation, the definition of a joule is used; 1 J = 1 kg m2 s-2)

  Deltam0 = DeltaE / c^2 = 890 kJ / (3.00 x 10^8 m s^-1)^2 = (890000 kg m^2 s^-2) / (9.00 x 10^16 m^2 s^-2) = 9.89 x 10^-12 kg = 9.89 ng  










The final answer shows that altogether, the products weigh 9.89 nanograms less than the reactants per mole of methane, much too insignificant to be observed by even the most precise balances. Therefore, the energy that was given off, the 890 kJ, came from 9.89 nanograms of matter.


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