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 (mass basis) of dissolved hydrogen is ......... Torr (rounded off to two decimal places). Given: For the hydrogen dissolution reaction in molten steel \( \left( \frac{1}{2} {H}_2(g) = [{H}] \right) \), the equilibrium constant (expressed in terms of ppm of dissolved H) is: \[ \log_{10} K_{eq} = \frac{1900}{T} + 2.4 \] 1 atm = 760 Torr. 
 

Hint:

When calculating equilibrium pressures for dissolved gases, make sure to apply the equilibrium constant formula correctly and ensure that you correctly convert units, particularly when working with concentrations expressed in ppm.

Answer 1

We need to find the equilibrium partial pressure of hydrogen at 1900 K. 
Step 1: Calculate \( K_{eq} \) at 1900 K Substituting \( T = 1900 \): \[ \log_{10} K_{eq} = \frac{1900}{1900} + 2.4 = 1 + 2.4 = 3.4 \] \[ K_{eq} = 10^{3.4} = 2511.88 \] Step 2: Use the equilibrium constant to calculate the partial pressure of hydrogen The equilibrium constant for the hydrogen dissolution reaction is: \[ K_{eq} = \frac{P_{{H}_2}}{[{H}]^2} \] Since 1 ppm corresponds to \( [{H}] = 10^{-6} \) (ppm is a mass basis, but we will assume concentration is proportional), we have: \[ P_{{H}_2} = K_{eq} \times [{H}]^2 = 2511.88 \times 10^{-6} = 0.0025 \, {Torr} \]