During the chlorination of NiO in a reactor at 900 K, it is required that 90% conversion of the chlorine gas be achieved during a single pass through the reactor. Calculate the required total gas pressure.

Question

Accepted Answer

The reaction is

[latex]NiO_{(s)}+Cl_{2{(g)}}=NiCl_{2(s)}+\frac{1}{2}O_{2(g)} [/latex]

for which [latex]\Delta G^{\circ }_{900 k}=-15,490 [/latex] J. Thus,

[latex]K_{900 K}=exp\left\lgroup\frac{15,490}{8.3144 imes 900} ight group =7.925 [/latex]

From the stoichiometry of the reaction,

[latex]NiO_{(s)}+Cl_{2{(g)}}=NiCl_{2(s)}+\frac{1}{2}O_{2(g)} [/latex]

For 90% conversion of the [latex]Cl_2,x=0.9[/latex], and thus, [latex]n_{Cl_2}=0.1,n_{O_2}=0.45[/latex] and [latex]n_T=0.55[/latex]. In a [latex]Cl_2-O_2[/latex] mixture at the pressure P , the partial pressures of chlorine and oxygen are thus

[latex]p_{Cl_2}=\frac{0.1}{0.55} P=0.182P[/latex] and [latex]p_{O_2}=\frac{0.45}{0.55} P=0.818P[/latex]

Thus,

[latex]K_{900 K}=7.925=\frac{p^{1/2}_{O_2} }{p_{Cl_2}}=\frac{(0.818P)^{1/2}}{0.182P} =\frac{4.969}{P^{1/2}} [/latex]

which has the solution P = 0.393 atm.
For another Quantitative Problem, see Appendix 12.B.

Question 12.11.3: During the chlorination of NiO in a reactor at 900 K, it is ...

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