help so. which ib the comrecz quiliertian corstast espession foe the tedlowing m
ID: 699002 • Letter: H
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so. which ib the comrecz quiliertian corstast espession foe the tedlowing maction (Hal Fe,0,10H,l 11.The reaction, 2 H2S (g) 2H2 (g)(g. wu armed on in a lolontainer and the following amounts were present it equilibrium: o.S0 M Hs, 1.0 M hh and 0.20 ws, wha is he value of the equilibrium constant, K? B.0.40 C O0D.13E1 12. Which of the following will resuls if some CHa (g) is removed Sron the equilibcum reactian A. i1:0 will be consamed B. More CHs and H0 will be produced C. Ke will decrease. D. More C0 will be produced. E. No change will occur 2 NO (g) + Bt; (g), (snom" 30 kinton if we For the reaction al equilibrium 2 NOBr (g) increase the reaction tempersture, A. more NOBr will be produced. B. NO will be consumed 13. C. Ke will decrease. D. Ke does not change. E. Ke will increase. 14. Consider the reaction 2 SO2 lg) + 02 (g)2 SO, (e), where K.-4.3x 10". If [S01-0010 M. [SO3] = 10 M and [O1-0010 M, which of the following statements most be true? A. The initial reaction mixture is already at equilibrium. B. ISO,] decreases, [SO2l and [Oal increase to reach equilibrium. C. [SOl increases, ISO-l and [O:] decrease to reach equilibrium. D. [SO,)- [sO] at equilibriun E. [SOI increases and [ decreases. 4 of 10Explanation / Answer
10.
A.Kc = [H2O]^3/ [H2]^3
Fe2O3(s) + 3H2 (g)--> 2Fe(s) + 3H2O(g)
The equilibrium constant expression is the ratio of the concentrations of the products over the reactants, raised to the power of its coefficient. Pure solids and liquids are not included in the equilibrium constant expression, because they do not affect the reactant amount at equilibrium in the reaction. So Fe2O3(s) and 2Fe(s) are exempted.
11)
C. 0.80
2H2S (g) ---->2H2(g) + S2(g) Kc(2)
At equilibrium [H2S] = 0.5M
[H2] = 1.0M
[S2]= 0.20M
Kc = [H2]^2 [S2]/ [H2S]^2
= [1.0]^2[0.20]/[0.5]^2
= 0.80
12)
B. More CH4(g) + H2O(g) will be produced.
CO(g) + 3H2(g) CH4(g) + H2O(g)
According to Le Chatelier's Principle, if we decrease the concentration of CH4, the position of equilibrium will move to the right to increase the concentration of CH4 and H2O again.
13)
E. Kc will increase.
2NOBr(g)-------->2NO(g) +Br2(g)
Enthalpy is positive, this means that a system absorbed heat and reaction is endothermic. If we increase the reaction temperature, more products are formed and Kc will increase.
Kc = [NO(g)]^2 [Br2(g)]/ [NOBr(g)]^2
14)
B. [SO3] will decreases and [SO2], [O2] increase to reach equilibrium.
2 SO2(g) + O2(g)---> 2 SO3(g)
Kc = 4.3 x 10^6
[SO2] = [O2] = 0.010 M
Q = reaction quotient = [SO3]^ 2 / [SO2]^ 2 [O2]
= (10)^2 / (0.010)^3
= 1.0 x 10^8 Since
Q > K, the equilibrium will shift to left, and [SO2] and [O2] will increase.
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