Consider the following system at equilibrium where AH° = 10.4 kJ, and K. = 1.80x10-2, at 698 K. %3D 2HI(g) H2(g) + I(g) When 0.23 moles of H2(g) are added to the equilibrium system at constant temperature: The value of K. The value of Q. The reaction must Orun in the forward direction to restablish equilibrium. Orun in the reverse direction to restablish equilibrium. remain the same. It is already at equilibrium. The concentration of I, will Submit Answer Retry Entire Group 4 more group attempts remaining <>

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Consider the following system at equilibrium where AH° = 10.4 kJ, and K. = 1.80×10, at 698 K.
2HI(g)
H2(g) + I(g)
When 0.23 moles of H2(g) are added to the equilibrium system at constant temperature:
The value of K.
The value of Q.
K.
The reaction must
Orun in the forward direction to restablish equilibrium.
run in the reverse direction to restablish equilibrium.
remain the same. It is already at equilibrium.
The concentration of I, will
Submit Answer
Retry Entire Group
4 more group attempts remaining
Transcribed Image Text:Consider the following system at equilibrium where AH° = 10.4 kJ, and K. = 1.80×10, at 698 K. 2HI(g) H2(g) + I(g) When 0.23 moles of H2(g) are added to the equilibrium system at constant temperature: The value of K. The value of Q. K. The reaction must Orun in the forward direction to restablish equilibrium. run in the reverse direction to restablish equilibrium. remain the same. It is already at equilibrium. The concentration of I, will Submit Answer Retry Entire Group 4 more group attempts remaining
Expert Solution
Step 1: Given

Given:

∆H° = 10.4 KJ

Kc = 1.80 × 10-2 

Tempreture = 698 K

Reaction:

2HI(g) <=> H2(g) + I2(g)

Moles of H2 added = 0.23 moles

Value of K= ?

Value of Q= ?

Reaction must = ?

Concentration of I= ?

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