4.0 moles of hydrogen iodide decompose to form hydrogen and iodine. If 0.442 mol
ID: 510445 • Letter: 4
Question
4.0 moles of hydrogen iodide decompose to form hydrogen and iodine. If 0.442 moles of iodine are detected after the reaction has reached equilibrium, what is the value of the equilibrium constant? Assume the reaction volume is 1.0 L. 2HI(g) H2(g) l2(g) (this reaction is reversible and undergoes equilbrium) A. 0.44 O B. 0,088 O C. 0.020 D. 1.4 x 10 QUESTION 3 4 points Save Answer The production of hydrogen chloride gas from hydrogen and chlorine has an equilibirum constant of 57.0. If 1.00 moles of hydrogen react with 1.00 moles of chlorine in a 10.0 L reaction flask, what is the equilibrium concentration of each reactant? H2(g) Cl2(g) 2HCl(g) (this reaction is reversible and undergoes equilbrium) A, 0.0791 M for both reactants B. 0.0210 M for both reactants O C. 0.136 M H2 O D. 0.0791 M H2 and 0.136 M Cl2Explanation / Answer
2
2HI --------------> H2 + I2
I 4 0 0
C -2*0.442 0.442 0.442
E 3.116 0.442 0.442
Kc = [H2][I2]/[HI]^2
Kc = 0.442*0.442/(3.116)^2 =0.02 >>>>>>>answer c
3.
H2 + Cl2 --------------> 2HCl
I 1 1 0
C -x -x +2x
E 1-x 1-x +2x
[H2] = 1-x/10
[Cl2] = 1-x/10
[HCl] = 2x/10
Kc = [HCl]^2/[H2][Cl2]
57 = (2x/10)^2/(1-x)/10*(1-x)/10
57 = (2x)^2/(1-x)^2
7.55 = 2x/1-x
7.55*(1-x) = 2x
x = 0.8
[H2] = 1-x/10 = 1-0.8/10 = 0.02M
[Cl2] = 1-x/10 = 1-0.8/10 = 0.02M >>>>>>answer b
4.
H2 + Cl2 --------------> 2HCl
I 0.1 0.2 0
C -x -x +2x
E 0.1-x 0.2-x +2x
Kc = [HCl]^2/[H2][Cl2]
57 = (2x)^2/(0.1-x)(0.2-x)
57*(0.1-x)(0.2-x) = 4x^2
x = 0.094
[HCl] = 2x = 2*0.094 = 0.189M>>>>>answer a
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