1.A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in
ID: 955875 • Letter: 1
Question
1.A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established:
2NO(g)+2H2(g)N2(g)+2H2O(g)
At equilibrium [NO]=0.062M.
a)Calculate the equilibrium concentration of H2.
b)Calculate the equilibrium concentration of N2.
c) Calculate the equilibrium concentration of H2O.
d)Calculate Kc.
2. At 218 C, Kc=1.2×104 for equilibrium NH4SH(s)NH3(g)+H2S(g)
a)Calculate the equilibrium concentration of NH3 if a sample of solid NH4SH is placed in a closed vessel at 218 C and decomposes until equilibrium is reached.
b) Calculate the equilibrium concentration of H2S if a sample of solid NH4SH is placed in a closed vessel at 218 C and decomposes until equilibrium is reached.
3.A mixture of hydrogen and nitrogen, which produces ammonia (NH3) in a reaction vessel, is allowed to reach equilibrium at a given temperature. The equilibrium mixture of gases contained 0.254 M NH3, 1.85 M N2, and 9.26 M H2. Calculate the equilibrium constant, K c at this temperature.
6)Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction
2COF2(g)CO2(g)+CF4(g), Kc=6.90
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
7)Consider the reaction
CO(g)+NH3(g)HCONH2(g), Kc=0.630
If a reaction vessel initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium?
Explanation / Answer
1.
NO= 0.1
H2 = 0.05
H2O = 0.1
2NO+2H2= N2+2H2O
K = N2*(H2O)^2 / (NO^2)(H2^2)
after reaction
NO= 0.1 -2x
H2 = 0.05-2x
H2O = 0.1 +2x
N2 = 0 +x
note that
[NO]=0.062M.
NO= 0.1 -2x = 0.062
x = (0.062-0.1)/(-2) = 0.019
then
NO= 0.1 -2*0.019 = 0.062
H2 = 0.05-2*0.019 = 0.012
H2O = 0.1 +2*0.019 = 0.138
N2 = 0 +x = 0.019
substitute in K
K = N2*(H2O)^2 / (NO^2)(H2^2)
K =(0.019)(0.138^2)/((0.062^2)(0.012^2))
K = 653.68
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