At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) + Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 x 10-2 atm, 2.0 × 10- 4 atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of H2?

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Chapter13: Fundamental Equilibrium Concepts
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At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) +
Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 × 10-2 atm, 2.0 × 10-
3.
4
atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of
6.
H2?
At 1000 K, Kp = 19.9 for the reaction Fe203(s) + 3 CO(g) = 2 Fe(s) + 3 CO2(g). What
is the value of K, for the reaction 4 Fe203(s) + 12 CO(g) = 8 Fe(s) + 12 CO2(g)?
Exactly 3.75 moles of N204 is placed in an empty 1.0-L container. After a period of
time, equilibrium is reached and described by the equation below:
N204 (g) = 2NO2 (g)
If at equilibrium the N204 is 45% dissociated, what is the value of the equilibrium
constant for the reaction?
Transcribed Image Text:At an elevated temperature, K, = 4.0 × 10-6 for the reaction 2 HBr(g) = H2(g) + Br2(g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 × 10-2 atm, 2.0 × 10- 3. 4 atm, and 2.0 × 10-4 atm, respectively, what is the equilibrium partial pressure of 6. H2? At 1000 K, Kp = 19.9 for the reaction Fe203(s) + 3 CO(g) = 2 Fe(s) + 3 CO2(g). What is the value of K, for the reaction 4 Fe203(s) + 12 CO(g) = 8 Fe(s) + 12 CO2(g)? Exactly 3.75 moles of N204 is placed in an empty 1.0-L container. After a period of time, equilibrium is reached and described by the equation below: N204 (g) = 2NO2 (g) If at equilibrium the N204 is 45% dissociated, what is the value of the equilibrium constant for the reaction?
What is the equilibrium equation for the reaction:
3
NH4NO3(s) = N20(g) + 2 H2O(/)?
O A) Ke = [N2O][H2O12
%3D
6.
5-
B) Kc = [N20][H2O]
8.
9.
--
N20]H20]2
Kc
%3D
NHẠNO3]
O D) K = [N20]
Transcribed Image Text:What is the equilibrium equation for the reaction: 3 NH4NO3(s) = N20(g) + 2 H2O(/)? O A) Ke = [N2O][H2O12 %3D 6. 5- B) Kc = [N20][H2O] 8. 9. -- N20]H20]2 Kc %3D NHẠNO3] O D) K = [N20]
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