Question

Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction. $CO(g)+H_{2}O(g)$⇄$CO_{2}(g)+H_{2}(g)$. If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that $P_{CO}$ = $P_{H2O}$ = 4.0 bar, what will be the partial pressure of $H_{2}$ at equilibrium ? Kp = 10.1 at 400°C.

Answer 1

The given chemical reaction is as follows.

$CO(g)\quad \quad +\quad { H }_{ 2 }O(g)\quad \quad \rightleftharpoons \quad C{ O }_{ 2 }(g)\quad \quad +\quad { H }_{ 2 }(g)$

$\begin{matrix} Initial\quad pressure \\ Final\quad pressure \end{matrix}\begin{matrix} 4.0 \\ 4.0-p \end{matrix}\begin{matrix} 4.0 \\ 4.0-p \end{matrix}\begin{matrix} 0 \\ p \end{matrix}\begin{matrix} 0 \\ p \end{matrix}$

${ K }_{ P }\quad =\quad \frac { { P }_{ C{ o }_{ 2 } }.{ P }_{ { H }_{ 2 } }\quad }{ { P }_{ CO }.{ P }_{ { H }_{ 2 }O } }$

$=\quad \frac { P.P }{ (4-P).(4-P) } \\ 10.1\quad =\quad \frac { { P }^{ 2 } }{ (4-P{ ) }^{ 2 } }$

$P\quad =\quad \frac { 12.71 }{ 4.17 } \quad =\quad 3.04\quad bar$

Therefore,$\quad { P }_{ { H }_{ 2 } }\quad =\quad 3.04\quad bar$