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Chapter 18: Problem 48

At temperatures between \(540^{\circ} \mathrm{C}(813 \mathrm{K})\) and \(727^{\circ} \mathrm{C}(1000 \mathrm{K}),\) the activation energy and preexponential for the diffusion coefficient of \(\mathrm{Na}^{+}\) in \(\mathrm{NaCl}\) are \(173,000 \mathrm{J} / \mathrm{mol}\) and \(4.0 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s},\) respectively. Compute the mobility for an \(\mathrm{Na}^{+}\) ion at \(600^{\circ} \mathrm{C}(873 \mathrm{K})\)

Short Answer

Expert verified
Answer: The mobility of the Na+ ion at 600°C (873 K) is approximately \(1.02 \times 10^{-8} \mathrm{m}^{2}/\mathrm{V} \cdot \mathrm{s}\).

Step by step solution

01

Understand the Arrhenius Equation

Arrhenius equation relates the diffusion coefficient (D) with temperature (T) using the activation energy (Ea) and the pre-exponential factor (D0). The equation is given by: \(D = D_{0} \mathrm{e}^{\frac{-Ea}{RT}}\) where: - D is the diffusion coefficient - \(D_{0}\) is the pre-exponential factor - Ea is the activation energy - R is the universal gas constant (8.314 J/mol⋅K) - T is the temperature in Kelvin Here, we are given the values for \(D_{0}\), Ea, and T. We will first use the Arrhenius equation to find the diffusion coefficient.
02

Calculate the Diffusion Coefficient

Plug in the given values of \(D_{0}\), Ea, and T into the Arrhenius equation: \(D = (4.0 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}) \mathrm{e}^{\frac{-(173,000 \mathrm{J} / \mathrm{mol})}{(8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K})(873 \mathrm{K})}}\) Now calculate the value of D: \(D \approx 3.39 \times 10^{-10} \mathrm{m}^{2}/\mathrm{s}\)
03

Understand the Nernst-Einstein Equation

Next, we will use the Nernst-Einstein equation to find the mobility (u) of Na+ ion. The equation is given by: \(u = \frac{D * q}{k_B * T}\) where: - u is the mobility - D is the diffusion coefficient - q is the charge of the ion (For Na+, q = 1.602 x 10⁻¹⁹ C) - \(k_B\) is Boltzmann constant (1.381 x 10⁻²³ J/K) - T is the temperature in Kelvin We have already calculated the value of D; therefore, we will use this value and the given temperature to find the mobility of the Na+ ion.
04

Calculate the Mobility of the Na+ Ion

Plug in the values of D, q, \(k_B\), and T into the Nernst-Einstein equation: \(u = \frac{(3.39 \times 10^{-10} \mathrm{m}^{2}/\mathrm{s})* (1.602 x 10^{-19} \mathrm{C})}{(1.381 x 10^{-23} \mathrm{J} / \mathrm{K}) * (873 \mathrm{K})}\) Now calculate the value of u: \(u \approx 1.02 \times 10^{-8} \mathrm{m}^{2}/\mathrm{V} \cdot \mathrm{s}\) So, the mobility of the Na+ ion at 600°C (873 K) is approximately \(1.02 \times 10^{-8} \mathrm{m}^{2}/\mathrm{V} \cdot \mathrm{s}\).

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