GENERAL CHEMISTRY II
CHEM 1312.001
EXAM 4
Friday May 1, 1998


1. If a chemical reaction is elementary, the exponents in its equilibrium constant expression will __________ agree with (match) the coefficients in the balanced equation.
a) always b) sometimes c) never
2. If a chemical reaction is non-elementary, the exponents in its equilibrium constant expression will __________ agree with (match) the coefficients in the balanced equation.
a) always b) sometimes c) never
3. A chemical equilibrium favors the reactants if

a) the numerical value of its equilibrium constant (Kc) is a very large (positive) number

b) the numerical value of its equilibrium constant (Kc) is a very small (positive) number

c) the numerical value of its equilibrium constant (Kc) is approximately equal to 1

d) the numerical value of its equilibrium constant (Kc) is a negative number

4. A chemical equilibrium favors the products if

a) the numerical value of its equilibrium constant (Kc) is a very large (positive) number

b) the numerical value of its equilibrium constant (Kc) is a very small (positive) number

c) the numerical value of its equilibrium constant (Kc) is approximately equal to 1

d) the numerical value of its equilibrium constant (Kc) is a negative number

5. At 25 oC, the reaction

     2SO2(g)      +      O2(g)      <---------->      2SO3(g)

has the equilibrium constant Kc = 8.0 x  1035. If a steel reaction vessel initially contained a stoichiometric mixture of SO2 and O2 at 25 oC and these gases were allowed to react to produce an equilibrium mixture, what would the composition of the equilibrium mixture be like?

a) Almost all of the SO2 and O2 would still be present. Only tiny traces of SO3 would be found in the vessel.

b) Very little SO2 and O2 would be present. There would be almost complete conversion to SO3.

c) An equal number of moles of SO2, O2, and SO3 would be present in the reaction vessel.

d) The compostion of the equilibrium mixture can not be determined, because the position of a chemical equilibrium is not related to the size of the equilibrium constant.

6. At 25 oC, the reaction

     2HF(g)      <---------->      H2(g)      +      F2(g)

has the equilibrium constant Kc = 1 x  10-95. If a steel reaction vessel initially contained only HF at 25 oC, and it was allowed to decompose to produce an equilibrium mixture, what would the composition of the equilibrium mixture be like?

a) Almost all of the original HF would still be in the reaction vessel. Only trace amounts of H2 and F2 would be present.

b) Only trace amounts of HF would remain in the vessel. Almost all of it would have decomposed to produce H2 and F2.

c) There would be an equal number of moles of HF, H2, and F2, in the reaction vessel.

d) The compostion of the equilibrium mixture can not be determined, because the position of a chemical equilibrium is not related to the size of the equilibrium constant.

7. At high temperatures, hydrogen iodide (HI) decomposes into the elements hydrogen (H2) and iodine (I2). In an experiment, a chemist heated HI gas to a temperature of 491 oC to decompose it. Not all of the the HI decomposed, because an equilibrium was established between HI, H2, and I2. The equation for the chemical equilibrium is

     2HI(g)      <---------->      H2(g)      +      I2(g)

The concentration of each gas in the equilibrium mixture was determined and the results were as follows:
[HI]eq = 0.3750 M [H2]eq = 0.0813 M [I2]eq = 0.0344 M
What is the numerical value of the equilibrium constant (Kc) for this reaction at 491 oC? Hint: This is a simple problem. It is not necessary to construct an ICE table, because all the equilibrium concentrations are given to you. Just plug them into the equilibrium constant expression for the reaction.
a) 1.06 x 10-3 b) 1.99 x 10-2 c) 50.3
d) 134 e) 953
8. Carbon monoxide (CO) and hydrogen (H2) react at high temperatures to produce methane (CH4) and water. The equilibrium equation for the reaction is

     CO(g)      +      3H2(g)      <---------->      CH4(g)      +      H2O(g)

In an experiment, a chemist put 0.1000 moles of CO and 0.3000 moles of H2 in a 1.00 L reaction vessel at 927 oC. No CH4 or H2O were initially present. After equilibrium had been established, 0.0387 moles of H2O was present in the reaction vessel. What is the numerical value of the equilibrium constant (Kc) for this reaction at 927 oC? Hints: Since you are given the equilibrium amount of only one of the substances in the equation, you must construct an ICE table to obtain the other concentrations. Note that the reaction must go to the right, since no products are initially present. It should be a simple task to fill in all the cells in the ICE table -- you don't need to solve an equation in terms of X. Fill in what you know already, then use stoichiometry and the fact that I + C = E to complete the table. Be aware that not all of the stoichiometric ratios are one to one in this equation. Look at the coefficients carefully. Once you have all the equilibrium concentrations, you can plug them into the equilibrium constant expression.
a) 0.133 b) 0.255 c) 0.734 d) 1.36 e) 3.93
9. Nitrogen (N2) and oxygen (O2) react at high temperatures to produce nitric oxide (NO). The equilibrium equation is

     N2(g)      +      O2(g)      <---------->      2NO(g)

At 2127 oC, the reaction has an equilibrium constant Kc = 2.5 x 10-3. Suppose the concentrations of N2 and O2 in an equilibrium mixture are as follows:
[N2]eq = 0.058 M [O2]eq = 0.027 M
What is the concentration of NO in this equilibrium mixture? Hint: Solve for the missing concentration in the equilibrium constant expression. Note that you can check the validity of your answer by plugging all your equilibrium concentrations back into the equilibrium constant expression. If your answer is correct, you should reproduce the original numerical value of the equilibrium constant.
a) 3.9 x 10-6 M b) 2.0 x 10-3 M c) 6.5 x 10-3 M
d) 4.1 x 10-2 M e) 3.8 x 10-1 M
10. At 727 oC, the reaction

     2SO2(g)      +      O2(g)      <---------->      2SO3(g)

has the equilibrium constant Kc = 4.17 x 10-2. A chemist started with a non-equilibrium mixture of SO2, O2, and SO3 at 727 oC in which the following initial concentrations existed:
[SO2]initial = 1.45 M [O2]initial = 2.83 M [SO3]initial = 0.11 M
What will be the direction of reaction as the system approaches equilibrium?

a) There will be reaction from left to right. (Net conversion of SO2 and O2 into SO3)

b) There will be reaction from right to left. (Net conversion of SO3 into SO2 and O2)

*** END OF TEST ***


ANSWERS:

1) a      2) a      3) b      4) a      5) b     

6) a      7) b      8) e      9) b      10) a     

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