Question

The number of vacancies in some hypothetical metal increases by a factor of 6 when the temperature is increased from 1070 K to 1160 K. Calculate the energy (in kJ/mol) for vacancy formation assuming that the density of the metal remains the same over this temperature range.

Answer 1

Answer:

The answer to the question is

The energy (in kJ/mol) for vacancy formation assuming that the density of the metal remains the same over this temperature range is 3.41165×10⁻²² kJ/mol

Explanation:

To solve the question, we note that the Boltzmann equation for vibrational states is

$\frac{N_{1} }{N_0} =e^{\frac{-dE}{kT} }$ Where

N₀ = Population of the lower energy state

N₁ = Population of the upper energy state

ΔE = dE = The energy difference between the two states

k = the Boltzmann constant and

T  = Kelvin temperature

Therefore we have

$6N_e ^{(\frac{-Q_v }{k(1070)})} = N_e ^{(\frac{-Q_v }{k(1160)}) }$

From which

$6 = \frac{N_e ^{(\frac{-Q_v }{k(1160)}) }}{N_e ^{(\frac{-Q_v }{k(1070)})}}$  

$6 = \frac{e ^{(\frac{-Q_v }{k(1160)}) }}{e ^{(\frac{-Q_v }{k(1070)})}}$

$6 = e ^{{(\frac{-Q_v }{k(1160)}) }{{+(\frac{Q_v }{k(1070)})}}}$

$ln(6) = {{(\frac{-Q_v }{k(1160)}) }{{+(\frac{Q_v }{k(1070)})}}}$

$Q_v =\frac{ ln(6) }{{(\frac{1 }{k(1160)}) }{{+(\frac{1 }{k(1070)})}}}$

Therefore

$Q_v$ = 3.41165×10⁻²² kJ/mol