Question

1. For the reaction 2A(g)⇌B(g)+2C(g), a reaction vessel initially contains only A at a pressure of...

1. For the reaction 2A(g)⇌B(g)+2C(g), a reaction vessel initially contains only A at a pressure of PA=265 mmHg. At equilibrium, PA=41 mmHg.

Calculate the value of Kp. (Assume no changes in volume or temperature.)

2. consider the following reaction:

A(g)⇌2B(g)

Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.

Kp= 2.0

PA, PB =

3. Consider the following reaction:
HC2H3O2(aq)+H2O(l)⇌H3O+(aq)+C2H3O−2(aq)
Kc=1.8×10−5 at 25∘C

If a solution initially contains 0.190 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25∘C?

4. Consider the following reaction:
Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)
A solution is made containing an initial [Fe3+] of 1.2×10−3 M and an initial [SCN−] of 7.9×10−4 M. At equilibrium, [FeSCN2+]= 1.7×10−4 M .

Calculate the value of the equilibrium constant (Kc).

5.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=4.00

If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

6.Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.890

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?

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Answer #1

1. Let us use the ICE table to calculate equilibrium pressure

2A B 2C
Initial 265 0 0
Change -2x +x +2x
Equilibrium 265-2x x 2x

Given that the equilibrium pressure of A = 41mmHg

265 - 2x = 41

x = 112

At equilibrium, p(A) = 41 mmHg

p(B) = x = 112 mmHg

p(C) = 2x = 224 mmHg

Kp = p(B)*p(C)^2/p(A)^2 = 112*224*224/41*41 = 3.343*10^3

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