chemistry Homework

Question

Radium-224 undergoes alpha decay with a half-life of 87.166 h. If we equate the energy released by the reaction with the enthalpy change and the amount of helium produced to the work, which is the free energy change, calculate the entropy change for the decay process for 1.00 mole of radium-224 against an external pressure of 1.000 bar. The masses of the isotopes are radium-224 = 224.020 211 8 u and radon-220 = 220.011 394 0 u.

Answer 1

For given Data:

Theory:

Radioactive decay involves the spontaneous transformation of one element into another. This can happen by changing the number of protons in the nucleus. In spontaneous decay, the relative energy and momentum of the parent molecule must be conserved. The final components of a given decay will be daughter nuclei and alpha particles.

Let,

x the mass of Parent Nuclie.

y be the mass of daughter nuclei

a is the mass of the alpha particle.

The Decay energy E=(Mx-My -Ma) c^2 as E= mc^2

E= (x-y - a) * 931.494 MeV/U

HERE,

x= 224.0202118 u

y = 220.0113940 u

a= 4.00203 u

E=( 224.0202118 u - 220.0113940 u - 4.00203 u) * 931.494 MeV/u

E= 6.323 MeV

The entropy change , ∆S= E/T = 6.323 MeV/ 298K = 1.01 * 10 ^-12 J /298k

∆S= 3.38 * 10 ^ -17 J/K