%3D Consider the following system at equilibrium where AH° = -10.4 kJ, and K. = 55.6, at 698 K. H2(g) + I(g) 2HI(g) If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of K. A. increases. B. decreases. C. remains the same. The value of Q. A. is greater than K. B. is equal to K C. is less than K.. The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. The number of moles of I, will: A. increase. B. decrease. C. remain the same.
%3D Consider the following system at equilibrium where AH° = -10.4 kJ, and K. = 55.6, at 698 K. H2(g) + I(g) 2HI(g) If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of K. A. increases. B. decreases. C. remains the same. The value of Q. A. is greater than K. B. is equal to K C. is less than K.. The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. The number of moles of I, will: A. increase. B. decrease. C. remain the same.
Chapter17: Spontaneity, Entropy, And Free Energy
Section: Chapter Questions
Problem 78E: Consider the following reaction at 298 K: 2SO2(g)+O2(g)2SO3(g) An equilibrium mixture contains O2(g)...
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Transcribed Image Text:Consider the following system at equilibrium where AH° = -10.4 kJ, and K. = 55.6, at 698 K.
H2(g) + I2(g)
2HI(g)
If the VOLUME of the equilibrium system is suddenly decreased at constant temperature:
The value of K.
A. increases.
B. decreases.
C. remains the same.
The value of Q.
A. is greater than K.
B. is equal to Ke
C. is less than K..
The reaction must:
A. run in the forward direction to reestablish equilibrium.
B. run in the reverse direction to reestablish equilibrium.
C. remain the same. It is already at equilibrium.
The number of moles of I, will:
A. increase.
B. decrease.
C. remain the same.
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Since equilibrium constant depends only on temprature not on volume so
The value of Kc remains same.
Since for the above reactions number of moles of gaseous species is same on both reactant and product side i.e ∆n = 0 so there will be no affect of changing volume on equilibrium.
Hence Qc remains same as that of Kc.
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