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Question

The vacancy concentration in a crystal doubles upon increasing the temperature from $27 \text{ } ^\circ\text{C}$ to $127 \text{ } ^\circ\text{C}$. The enthalpy (in $kJ \text{ mol}^{-1}$) of vacancy formation is: _________ (round off to 2 decimal places).
Given: $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$

Understanding Vacancy Concentration and Temperature

Vacancy concentration ($n_v$) in crystals increases with temperature ($T$). This dependence follows an Arrhenius-like relationship:

$n_v \propto e^{-E_f / (RT)}$

where $E_f$ is the enthalpy of vacancy formation and $R$ is the ideal gas constant.

Calculating Enthalpy of Vacancy Formation ($E_f$)

Given: Initial temperature $T_1 = 27 \text{ } ^\circ\text{C}$, final temperature $T_2 = 127 \text{ } ^\circ\text{C}$. The vacancy concentration doubles ($n_{v2} = 2 n_{v1}$). $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$.

  1. Convert Temperatures to Kelvin:
    • $T_1 = 27 + 273.15 = 300.15 \text{ K}$
    • $T_2 = 127 + 273.15 = 400.15 \text{ K}$
  2. Formulate Concentration Ratio: The ratio of vacancy concentrations relates to the absolute temperatures: $ \frac{n_{v2}}{n_{v1}} = e^{E_f \left(\frac{1}{T_1} - \frac{1}{T_2}\right)} $ Substitute the given doubling factor ($n_{v2} / n_{v1} = 2$): $ 2 = e^{E_f \left(\frac{1}{300.15 \text{ K}} - \frac{1}{400.15 \text{ K}}\right)} $
  3. Solve for $E_f$: Apply the natural logarithm to both sides: $ \ln(2) = E_f \left(\frac{1}{300.15 \text{ K}} - \frac{1}{400.15 \text{ K}}\right) $ Calculate the difference in inverse temperatures: $ \Delta\left(\frac{1}{T}\right) = \frac{1}{300.15 \text{ K}} - \frac{1}{400.15 \text{ K}} \approx 0.00083258 \text{ K}^{-1} $ Calculate $E_f$ in J/mol: $ E_f = \frac{\ln(2)}{\Delta\left(\frac{1}{T}\right)} \approx \frac{0.693147}{0.00083258 \text{ K}^{-1}} \approx 8325.9 \text{ J mol}^{-1} $
  4. Final Result: Convert $E_f$ to kJ/mol and round to two decimal places: $ E_f \approx \frac{8325.9}{1000} \text{ kJ mol}^{-1} \approx 8.33 \text{ kJ mol}^{-1} $ The calculated enthalpy of vacancy formation is $8.33 \text{ kJ mol}^{-1}$.
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Important Questions from Diffusion Fick's Second Law Concentration Profile

  1. During carburizing of a steel, the surface concentration is kept constant at 1.4 wt.% carbon. Diffusivity of carbon for the steel at 950 $^\circ$C is $6.25 \times 10^{-11}$ m$^2$/s. At 950 $^\circ$C, the time required to carburize the steel with an initial composition of 0.2 wt.% carbon to 0.8859 wt.% carbon at a depth of 0.2 mm is ______________ seconds (approximate to the nearest integer).

     Use the nearest value of the error function from the table given below for your calculation.

    zerf (z)
    0.30.3268
    0.40.4284
    0.50.5205
  2. What is the depth (in $µm$) from the surface of the specimen at which a composition of 0.4 wt.% C is obtained after carburizing at $870^\circ C$ for 10 h?
  3. For self-diffusion in polycrystalline copper with a lattice diffusion coefficient $D_L$, grain boundary diffusion coefficient $D_{GB}$, and surface diffusion coefficient $D_S$, the correct relationship is

  4. The concentration $C$ of a solute (in units of atoms$\cdot\text{mm}^{-3}$) in a solid along $x$direction (for $x > 0$) follows the expression
    $C = a_1x^2 + a_2x$
    where $x$ is in mm, $a_1$ and $a_2$ are in units of atoms$\cdot\text{mm}^{-5}$ and atoms$\cdot\text{mm}^{-4}$,respectively. Assuming $a_1= a_2= 1$, the magnitude of flux at $x = 2 \text{ mm}$ is________ $\times 10^{-3} \text{ atoms} \cdot \text{mm}^{-2} \cdot \text{s}^{-1}$ (answer rounded off to the nearest integer).
    Given: diffusion coefficient of the solute in the solid is $3 \times 10^{-3} \text{ mm}^2 \cdot \text{s}^{-1}$.
  5. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]
    Assertion [a]: The rate of homogenization in a dilute substitutional solid solution of B in A is controlled by the diffusivity of B.
    Reason [r]: Atomic migration cannot occur along dislocations and grain boundaries.
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