Question

Consider the methanol decomposition into carbon monoxide and hydrogen according to theequilibrium reaction below.CH3 OH(g)>CO (g)+ 2 H2(g)The equilibrium constant for the reaction is K = 0.04. You let the reaction progressinto a 10.0L balloon and patiently wait for equilibrium to be reached. Onceequilibrium reached, you observe that the balloon contains 1.38mol of CO and 1.26mol ofH2a) What is the expression allowing you to calculate the equilibrium constant of thisreaction from the concentrations?b) At equilibrium, the flask contains how many moles of CH3OH?c) You want to upset the balance to the right. Should you add or removeCO in the balloon? Briefly justify your choice using the principle of LeChatelier.

Answer 1

`CH_3OH_((g))\rightleftarrows CO_((g))+2H_(2(g))`

A) Equilibrium constant is usually written as:

`\begin{gathered} aA\rightarrow bB\text{ } \\ k=([B]^b)/([A]^a) \\ \\ K=([CO]*[H2]^2)/([CH3OH]) \end{gathered}`

We will firstly determine the concentrations of the gases:

`\begin{gathered} c=(n)/(V) \\ c:concentration \\ n:moles \\ V:volume \\ \text{ }[CO]=(1.38)/(10.L)=0.138M \\ \\ \text{ }[H2]=(1.26)/(10.0L)=0.126M \end{gathered}`

B) By substituting what we know in the first equation to find out our unknown we have:

`\begin{gathered} 0.04=\frac{[{0.138][0.126]^2}}{x} \\ x=0.055M \\ \end{gathered}`

To determine the number of moles of CH3OH, we multiply the molar concentration by 10L

`\begin{gathered} _nCH3OH=0.055*10 \\ _nCH3OH=0.55mol \end{gathered}`

At equilibrium we have 0.55mol of CH3OH

C) If you want the equilibrium to to be shifted to the right in the direction of the product you can remove the CO from the ballon. According to Le Chateliers principle, the foward reaction is favored if the concentration of the the product is deecreased so that more product can be formed.