.. Consider the following system at equilibrium where AHP-111 kJ/mol, and Kc630,
ID: 1019336 • Letter: #
Question
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Consider the following system at equilibrium where AHP-111 kJ/mol, and Kc630, at 723 K 2 NH3 (g) N2 (g) +3 H2 (g) When 0.29 moles of NH3 (g) are removed from the equilibrium system at constant temperature: the value of K, A. increases. B. decreases. C. remains the same. A. is greater tharn B. is equal to K. C. is less than Kc the value of A. run in the forward direction to reestablish equilibrium B. run in the reverse direction to reestablish equilibrium C. remain the same. It is alreadv at equilibrium the reaction must: the concentration of N2 will: A. increase. B. decrease. C. remain the same.Explanation / Answer
Question 1.
Value of Kc changes when there is change in Temperature. So, as the temperature is constant, Kc remains the same.
For Qc, we assume that the initial number of moles to be 1.
So, number of moles of:
NH3= 1 - 0.29
N2= 0.29/2
H2= 3*0.29/2
This is according to the number of moles present in the equation.
So, Qc = [N2 * (H2)^3 ] / (NH3)^2 = 0.0236
Qc is lesser than Kc
As Qc < Kc , the reaction has to move forward to maintain equilibrium. So, concentration of N2 will increase.
Question 2 :
Rate Law = k. [OCl-]^1 . [I-]^x.
As, it is overall second order, x = 1.
rate law = k [OCl-] [I-].
Question 3:
Rate = k. [UO2]^x . [H]^y
So for instance 1 ad 2,
1.16*10^-4 = k. [1.81*10^-3]^x . [0.210]^y
4.65*10^-4 = k. [3.62*10^-3]^x . [0.210]^y.
Dividing second by first:
4 = 2^x
x = 2.
For instance 4,
9.30*10^-4 = k. [3.62*10^-3]^x . [0.420]^y
Putting this x = 2 in instance 4 and 2,
9.30*10^-4 = k. [3.62*10^-3]^2 . [0.420]^y -------- 1
4.65*10^-4 = k. [3.62*10^-3]^2 . [0.210]^y -------- 2.
Dividing 1 by 2,
2 = 2^y
y = 1.
So, rate Law = k. [UO2]^2 . [H]^1.
By putting the values for x and y in any of the instances, we can get the value for k.
I got the value of k to be 168.97.
Let me know if there is ny issue related to this.
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