Investigate the effect of temperature on the disproportionation of toulene, give

Question

Investigate the effect of temperature on the disproportionation of toulene, given the data:

T (Celsius) 50 100 150 200 250

K(ortho-) .059 .065 .07 .074 .078

K(meta-) .206 .208 .209 .21 .211

K(para-) .0837 .0875 .0904 .0928 .0949

K(otho-),K(meta-),K(-para-) are the thermodynamic equilibrium constants for the disproportionation reaction yielding benzene and ortho-,meta-,and para-xylene respectively. What temperature would you choose to effect (a) the best conversion to meta-sylene, and (b) the best conversion to benzene. What effect would total pressure have on the product distribution?

Explanation / Answer

Consider the reaction for the formation of meta-xylene :

2 toulene (g) ---> benzene (g) + m-xylene (g)

(A)

As we can see from the data provided, the maximum value of K(meta-) appears for the temperature 250 'C. So out of the given temperatures, 250 'C is the best option to get maximal m-xylene.

(B)

Same as part (A), because the products are formed in a 1:1 ratio, so the temperature at which we get maximal xylene will also be the temperature when we get maximal benzene.

Since the total number of gaseous moles is the same on both the reactant as well as the product side, total pressure exhibits no effect on the product distribution.

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