Question

Ammonia is an important raw material used in the manufacture of, among other things, fertilizers. Ammonia can be produced by the catalytic reaction between nitrogen and hydrogen. The overall reaction is Industrially, this reaction is carried out at elevated pressures where the gases are not ideal. Calculate the equilibrium composition for this operation at a reac- tor pressure of 1000 bar and temperature of 723 K. The feed consists of 21 mol% N 2, 63 mol% H2, and 16 mol% inert Ar. The fugacities of pure hydrogen, nitro- gen, and ammonia are 1350, 1380, and 860 bar, respectively. Other thermody- namic data are available in Appendix 1.

Answer 1

Basis: 1 mole of feed gases

reaction: $\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightleftharpoons NH_{3}$

   $N_{2}+3H_{2}\rightleftharpoons 2NH_{3}$

From thermodynamic data, K_p=4.5 $\times$ 10^-5 at 723 K.

now, $K_{c}={K_{p}}.P^{-\nu }$

$K_{p}=K_{c}.P^{\nu }$ ..........................................( $\bullet$ )

now, $K_{c}=\frac{y_{NH_{3}}^{2}\phi _{NH_{3}}^{2}}{y_{N_{2}}^{1}\phi _{N_{2}}.y_{H_{2}}^{3}\phi _{H_{2}}^{3}}$ ...............................................................( $\ast$ )

$\phi _{i}$ = Fugacity of species i

calculation of the mole fractions are done by the following relation,

$y_{i}=\frac{n_{i}}{n}=\frac{n_{io}+\nu _{i}\varepsilon }{n_{o}+\nu \varepsilon }$

where, n_io=initial moles of species i

n_o= total moles before reaction

$\varepsilon$ = extent of reaction

$\nu$ = net stoichiometric co-efficient of the reaction $(\nu =\sum \nu _{i} )$

$\nu _{i}$ = stoichiometric co-efficient of species i, (+ve for products, -ve for reactants)

$\nu$ =2-3-1= -2

$\therefore y_{N_{2}}=\frac{0.21-\varepsilon }{1+(-2)\varepsilon }$

$y_{H_{2}}=\frac{0.63-3\varepsilon }{1-2\varepsilon }$

$y_{NH_{3}}=\frac{0+2\varepsilon }{1-2\varepsilon }$

now,( $\ast$ )=> $K_{c}=\frac{(\frac{2\varepsilon }{1-2\varepsilon })^{2}.860^{2}}{(\frac{0.21-\varepsilon }{1-2\varepsilon })^{1}.1380^{^{}}\times (\frac{0.63-3\varepsilon }{1-2\varepsilon })^{3}.1350^{3}}$

so, ( $\bullet$ )=> $K_{p}=\frac{(\frac{2\varepsilon }{1-2\varepsilon })^{2}.860^{2}}{(\frac{0.21-\varepsilon }{1-2\varepsilon })^{1}.1380\times (\frac{0.63-3\varepsilon }{1-2\varepsilon })^{3}.1350^{3}}\times P^{-2}$

putting , values for K_p, P=1000 bar, then solving we will get,

$\varepsilon$ =0.015

Mole fractions at equilirium

$\therefore y_{N_{2}}=\frac{0.21-0.015 }{1+(-2)\times 0.015 }=0.201$

$y_{H_{2}}=\frac{0.63-3\times 0.015}{1-2\times 0.015}=0.603$

$y_{NH_{3}}=\frac{2\times 0.015}{1-2\times 0.015}=0.030$

$y_{inert}=(1-0.201-0.603-0.030)=0.166$