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question

Determine the equilibrium constant for the dissociation of sodium at $298 \mathrm{~K}: \mathrm{Na}_2(g) \rightleftharpoons 2 \mathrm{Na}(g)$. For Na2\mathrm{Na}_2, $B=0.155 \mathrm{~cm}^{-1}, \widetilde{\nu}=159 \mathrm{~cm}^{-1}$, the dissociation energy is $70.4 \mathrm{~kJ} / \mathrm{mol}$, and the ground-state electronic degeneracy for Na\mathrm{Na} is 2 .
Plain-text mathematical notation (without MathML)
Determine the equilibrium constant for the dissociation of sodium at $298 \mathrm{~K}: \mathrm{Na}_2(g) \rightleftharpoons 2 \mathrm{Na}(g)$. For Na₂, $B=0.155 \mathrm{~cm}^{-1}, \widetilde{\nu}=159 \mathrm{~cm}^{-1}$, the dissociation energy is $70.4 \mathrm{~kJ} / \mathrm{mol}$, and the ground-state electronic degeneracy for Na is 2 .
Original LaTeX notation
Determine the equilibrium constant for the dissociation of sodium at $298 \mathrm{~K}: \mathrm{Na}_2(g) \rightleftharpoons 2 \mathrm{Na}(g)$. For $\mathrm{Na}_2$, $B=0.155 \mathrm{~cm}^{-1}, \widetilde{\nu}=159 \mathrm{~cm}^{-1}$, the dissociation energy is $70.4 \mathrm{~kJ} / \mathrm{mol}$, and the ground-state electronic degeneracy for $\mathrm{Na}$ is 2 .

category

physics

options

  1. 2.25 $10^{-9}$
  2. 1.98 $10^{-9}$
  3. 5.00 $10^{-8}$
  4. 4.50 $10^{-9}$
  5. 3.15 $10^{-9}$
  6. 1.75 $10^{-9}$
  7. 7.80 $10^{-9}$
  8. 2.75 $10^{-9}$
  9. 3.00 $10^{-9}$
  10. 1.10 $10^{-10}$

src

scibench-thermo

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