Q36E A necessary step in the manufact... [FREE SOLUTION]

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Chapter 13: Q36E (page 753)

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide (\({\rm{S}}{{\rm{O}}_3}\)), from sulfur dioxide (\({\rm{S}}{{\rm{O}}_2}\)), and oxygen (\({{\rm{O}}_2}\)), shown here.
\(2{\text{S}}{{\text{O}}_2}(g) + {{\text{O}}_2}(g) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}(g)\)
At high temperatures, the rate of formation of \({\rm{S}}{{\rm{O}}_3}\)is higher, but the equilibrium amount (concentration or partial pressure) of \({\rm{S}}{{\rm{O}}_3}\) is lower than it would be at lower temperatures.
(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?
(b) Is the reaction endothermic or exothermic?

Short Answer

Expert verified

a) The equilibrium constant will decrease.

b) The reaction is exothermic.

Step by step solution

Calculating increase or decrease in equilibrium constant

The reaction is

\(2{\text{S}}{{\text{O}}_2}({\text{g}}) + {{\text{O}}_2}({\text{g}}) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}({\text{g}})\)

At high temperatures, the equilibrium amount of \(\left[ {{\rm{S}}{{\rm{O}}_3}} \right]\)is lower than it would be at lower temperatures.

(a)

Let us see what will happen with the equilibrium constant if the temperature is increased.

The equilibrium constant is

\({K_c} = \frac{{{{\left[ {{\rm{S}}{{\rm{O}}_3}} \right]}^2}}}{{{{\left[ {{\rm{S}}{{\rm{O}}_2}} \right]}^2} \cdot \left[ {{{\rm{O}}_2}} \right]}}\)

The increase in temperature will decrease the concentration of \(\left[ {{\rm{S}}{{\rm{O}}_3}} \right]\), and increase the concentration of \(\left[ {{\rm{S}}{{\rm{O}}_2}} \right]\)and \(\left[ {{{\rm{O}}_2}} \right]\), therefore, the value of equilibrium constant will decrease.

Determine if reaction endothermic or exothermic

(b) Let us assume that the reaction is endothermic.

\(2{\text{S}}{{\text{O}}_2}({\text{g}}) + {{\text{O}}_2}({\text{g}}) + {\text{heat}} \rightleftharpoons 2{\text{S}}{{\text{O}}_3}({\text{g}})\)

Increase in temperature will shift the equilibrium to the right, resulting increase in concentration of \(\left[ {{\rm{S}}{{\rm{O}}_3}} \right]\) and decrease in concentration of \(\left[ {{\rm{S}}{{\rm{O}}_2}} \right]\)and \(\left[ {{{\rm{O}}_2}} \right]\).

Let us assume that the reaction is exothermic.

\(2{\text{S}}{{\text{O}}_2}({\text{g}}) + {{\text{O}}_2}({\text{g}}) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}({\text{g}}) + {\text{heat}}\)

The increase in temperature will shift the equilibrium to the left, resulting in decrease in concentration of \(\left[ {{\rm{S}}{{\rm{O}}_3}} \right]\)and increase in concentration of \(\left[ {{\rm{S}}{{\rm{O}}_2}} \right]\)and \(\left[ {{{\rm{O}}_2}} \right]\).

Therefore, the given reaction is exothermic.