Question

use the references to access important values if needed for this question. consider the following system at equilibrium where \\( \delta h ^ { \prime } = - 198 \mathrm { kj } \\), and \\( k _ { c } = 34.5 \\), at 1150 k: \\( 2 \mathrm { so } _ { 2 } ( g ) + \mathrm { o } _ { 2 } ( g ) \
ightleftharpoons 2 \mathrm { so } _ { 3 } ( g ) \\) when 0.16 moles of \\( \mathrm { so } _ { 3 } ( g ) \\) are removed from the equilibrium system at constant temperature: the value of \\( k _ { c } \\) increases decreases remains the same the value of \\( q _ { c } \\) is greater than \\( k _ { c } \\) is equal to \\( k _ { c } \\) is less than \\( k _ { c } \\) the reaction must run in the forward direction to reestablish equilibrium run in the reverse direction to reestablish equilibrium remain in the current position, since it is already at equilibrium the concentration of \\( \mathrm { so } _ { 2 } \\) will increase decrease remain the same

Explanation:

• For \(K_{c}\): The equilibrium constant \(K_{c}\) only depends on temperature. Since the temperature is constant, \(K_{c}\) remains the same.
• For \(Q_{c}\): The reaction quotient \(Q_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\). When \(SO_{3}\) is removed (\([SO_{3}]\) decreases), \(Q_{c}\) becomes less than \(K_{c}\).
• For the reaction direction: When \(Q_{c}

ightleftharpoons 2SO_{3}(g)\)) to re - establish equilibrium.

• For \([SO_{2}]\): As the reaction runs in the forward direction, \(SO_{2}\) is consumed, so its concentration decreases.

Answer:

• The value of \(K_{c}\): remains the same
• The value of \(Q_{c}\): is less than \(K_{c}\)
• The reaction must: run in the forward direction to reestablish equilibrium
• The concentration of \(SO_{2}\) will: decrease