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Consider the following system at equilibrium where ?11° 198 kJ, and Kc-34.5 , at

ID: 1040731 • Letter: C

Question

Consider the following system at equilibrium where ?11° 198 kJ, and Kc-34.5 , at 1.15x 103 K: 2502 (g) + O2 (g)--2SO3 (g) If the TEMPERATURE on theequibium system is suddenly decreased The value of KcA. Increases B. Decreases C. Remains the same A. Is greater than Kc B. Is equal to Kc C. Is less than Kc The value of Qc The reaction must:A. Run in the forward direction to restablish equilibrium. B. Run in the reverse direction to restablish equilibrium. C. Remain the same. Already at equilibrium. The concentration of O2 will:A. Increase. B. Decrease C. Remain the same.

Explanation / Answer

The value of ?H is negative for the given reaction, this means the reaction is exothermic.

Now, according to Le Chateliers principle

Decrease in temperature will favour the reaction in a direction of exothermic side. Since forward reaction is exothermic, so reaction will proceed in forward reaction.

If the temperature is suddenly decreased then the value of Kc is not affected immediately. It will take some time to re- establish the equilibrium. So, the correct option is

# C remains the same

Qc is the reaction quotient. It gives the value at a state other than state of equilibrium. As explained above the decrease in temperature will favour the forward reaction, so more product will be formed. Hence , the correct option is

#A Qc will be greater than Kc

* Kc = [SO3]^2/([SO2]^2 ×[O2])

For Qc the expression will be similar to Kc but the concentration value will be other than the equilibrium concentration.

The expression will have greater value of products concentration when temperature is decreased, so Qc will be greater.

As explained above it is clear that the correct option is

# A the reaction will proceed in forward direction to re establish equilibrium.

Since the reaction will proceed in forward direction, so more product Will be formed and concentration of reactants will decrease. So, the correct option is

# B Decrease