Question

Molten steel at 1900 K having dissolved hydrogen needs to be vacuum degassed. The equilibrium partial pressure of hydrogen to be maintained to achieve 1 ppm of dissolved hydrogen is .............. Torr (rounded off to two decimal places).

Hint:

Remember: ppm of solute is taken as concentration for equilibrium expressions. Always square the ratio when the stoichiometry involves one-half mole of H$_2$.

Answer 1

Step 1: Write the given equation.
\[ \log_{10} K_{eq} = -\frac{1900}{T} + 2.4 \] At $T = 1900$ K: \[ \log_{10} K_{eq} = -\frac{1900}{1900} + 2.4 = -1 + 2.4 = 1.4 \] \[ K_{eq} = 10^{1.4} \approx 25.1 \] Step 2: Definition of equilibrium constant.
For half H$_2$(g) equal to [H], \[ K_{eq} = \frac{[H]}{(p_{H2})^{1/2}} \] Rearrange for $p_{H2}$: \[ p_{H2} = \left(\frac{[H]}{K_{eq}}\right)^2 \] Step 3: Substitute values.
For 1 ppm of dissolved H, $[H] = 1$: \[ p_{H2} = \left(\frac{1}{25.1}\right)^2 = 0.00159 \, \text{atm} \] Convert to Torr: \[ p_{H2} = 0.00159 \times 760 = 1.21 \, \text{Torr} \] Rounded to two decimals: \[ \boxed{\text{Equilibrium pressure = 1.21 Torr}} \]