Question How do you set up the equations needed to solve the chemical equilibrium of methane steam reforming using the law of mass action and the reactions stoichiometry? How the equilibrium constant of the reactions changes with temperature. What are the main characteristics of this method to solve chemical equilibrium compared to non-stoichiometric methods such as the Lagrange Multiplier method?

Solution :- Step 1: Finding equilibrium constantThe reaction for methane steam reforming is:\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O}<==>\mathrm{CO}+3 \mathrm{H}_{2}This is how we can use law of mass action and stoichiometry to find out the equilibrium constant:according to law of mass action:rate of forward reaction =r_{f}=k_{f}\left(P_{\text {methane }}\right)rate of forward reaction =r_{b}=k_{b}\left(P_{C O}\right)\left(P_{h y d}\right)^{3}for equilibrium rate of forward reaction = rate of backward reaction==r_{f}=r_{b}==k_{f}\left(P_{\text {methane }}\right)=k_{b}\left(P_{\mathrm{CO}}\right)\left(P_{\text {hyd }}\right)^{3}==>k_{f} / k_{b}=\left(P_{\mathrm{CO}}\right)\left(\mathrm{P}_{\mathrm{hyd}}\right)^{3} /\left(\mathrm{P}_{\text {methane }}\right)= equilibrium constant =\mathrm{K}_{\mathrm{eq}}Step 2: Change with Temperature &amp; Characteristics of this methodSince this is an endothermic reaction, on increasing temperature, the forward reaction become faster than backward reaction. And on decreasing temperature, the reverse happens, i.e, backward reaction becomes more prominent. This is why methane steam reforming is carried out at a very high temperature, around 700-1000 degrees C.Main characteristics of this method is that it tries to measure rate of forward and backward reactions using stoichiometry of the reaction (so, a balanced equation is required). Equilibrium is assumed to be that point when rates of backward and forward reactions become equal. Langrange multiplier method instead looks at the thermodynamics of the system and tries to find that point where Gibbs free energy is minimum. &#160; &#160; &#160; ...