Chapter 16 1. A) True B) False 2. Reaction rates can be varied by changing eithe
ID: 997314 • Letter: C
Question
Chapter 16
1.
A) True
B) False
2.
Reaction rates can be varied by changing either the concentrations of the reacting species or the temperature at which the reaction is carried out.
A) True
B) False
3.
If a chemical reaction at equilibrium is disturbed by increasing the amount of products present, a left-to-right equilibrium shift (reactants becoming products) is necessary to restore an equilibrium condition.
A) True
B) False
4.
At equilibrium, the concentrations of reactants and products are always exactly the same.
A) True
B) False
5.
Of the two reactions whose equilibrium constants are 0.003 and 30, the one which will produce greater quantities of products is the one whose equilibrium constant is 30.
A) True
B) False
6.
A) True
B) False
7.
For chemical reactions to occur at all, "old" bonds must be broken and "new" bonds must form.
A) True
B) False
8.
An "equilibrium condition" can be achieved only when the reactant-to-product conversion is complete. (100% of the reactants have been converted to products.)
A) True
B) False
9.
In the reaction Cl2 (g) + 3F2 (g) 2ClF3 (g), reducing the pressure will shift the equilibrium to the right.
A) True
B) False
10.
A reaction which favors products has Keq < 1.
A) True
B) False
11.
Predict the effect of increasing pressure in each of the reactions in the first column by choosing the appropriate direction shift in the right column.
ABCl2 (g) + H2 (g) 2 HI (g)
ABCO2 (g) + 2F2 (g) 2OF2 (g)
ABCCH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
12.
Consider the reaction: C(s) + CO2 (g) 2 CO (g) (endothermic). Match each equilibrium change described in the left column with the corresponding consequence in the right column.
ABCadding heat
ABCadding a catalyst
ABCadding carbon monoxide
ABCadding extra carbon
ABCadding extra carbon dioxide
ABCincreasing the pressure
ABCcooling the mixture
13.
Match each reaction and its equilibrium concentrations that appear in the left column with the corresponding equilibrium constant value (Kp) that appears in the right column.
14.
Identify each of the equilibrium constant values in the first column as "Equilibrium lies to the right/products favored." or "Equilibrium lies to the left/reactants favored."
AB0.00563
AB400.23
AB2.3 x 103
AB4.5 x 10-3
AB3.7 x 103
AB2356.3
15.
Predict the effect of increasing the concentration of the bolded species in each of the balanced equations in the first column by choosing the appropriate direction shift in the right column.
AB2NH3 (g) N2 (g) + 3H2 (g)
ABl2 (g) + H2 (g) 2 HI (g)
ABO2 (g) + 2F2 (g) 2 OF2 (g)
Chapter 17
16.
Alcohols are the only organic compounds that can hydrogen bond.
A) True
B) False
17.
Specific enzymes are needed for the digestion of proteins, carbohydrates, and fats.
A) True
B) False
18.
Without enzymes, high temperatures and extreme pH would be necessary to hydrolyze carbohydrates, proteins, and fats quickly.
A) True
B) False
19.
The condensed notation for an ester is ROR.
A) True
B) False
20.
Protein molecules are complex such that a complete picture of each includes a description of its primary, secondary, and tertiary structures.
A) True
B) False
Explanation / Answer
1. (B) False: At equilibrium, the rates of forward (consumption of reactants) and backward (consumption of products) reactions are equal.
2. (A) True: The rate of a reaction depends on the concentration of reactant(s). The activation energy of a reaction is temperature dependent and the reaction rate depends on the activation energy.
3. (B) False: The explanation can be found here. Let us consider a hypothetical reaction
aA + bB <===> cC + dD, the equilibrium constant for this reaction (the reaction temperature is kept constant) is
K =[C]c[D]d/[A]a[B]b
K is constant at a particular temperature. If the reaction is disturbed by adding more products, the numerator of the expression increases. To keep K constant, the denominator must increase proportionally. Hence, products are converted into reactants. In other words, a right-to-left shift is necessary.
4. (B) False: Look at the expression for K above. At a particular temperature, K must remain constant which means that the ratio of the products and the reactants must remain constant; the concentration of reactants and products need not be equal, but the ratio must be constant.
5. (A) True: Again, consider the expression for the equilibrium constant from (3) above. Two reactions have values of K as 0.003 and 30; the reaction having value of K as 30 will have the numerator higher than the denominator, i.e., more products must be formed.
6. (B) False For gaseous reactions, the equilibrium constant can be written in terms of partial pressures at equilibrium. For the given reaction
N2H4 (g) <====> 2 H2 (g) + N2 (g)
Kp = (pH2)2(pN2)/(pN2H4) = (0.20 atm)2(0.10 atm)/(0.10 atm) = 0.04
7. (A) True: Let us look at a simple example – 2 H2 + O2 -----> 2 H2O
Bonds between H atoms in H2 molecule and O atoms in O2 molecule are broken and new O-H bonds in H2O are formed.
8. (B) False: At 100% conversion, the reaction is irreversible. Again, consider the equilibrium expression from (3). If 100% conversion occurs, the denominator becomes non-existent and hence we cannot write the equilibrium expression.
9. (B) False: The equilibrium expression in terms of partial pressures is
Cl2 (g) + 3 F2 (g) <===> 2 ClF3 (g)
Kp = (pC)2/(pA)(pB)3 where we denote Cl2 , F2 and ClF3 as A,B and C.
Partial pressures are related to total pressure P as
pA = xA.P
pB = xB.P
pC = xC.P where the x terms denote mole fractions
Hence, Kp = (xc.P)2/(xA.P)(xB.P)3 = (xc2/xA.xB3).(1/P2)
When the pressure is reduced, the denominator of the expression decreases. Kp is constant as temperature is constant. To achieve this, the numerator must decrease proportionally. Thus, less products will be formed, i.e, the left hand reaction is favoured.
10. (B) False: Consider the expression for K from (3) above. A reaction that favours more products should have the numerator (in the expression for K) higher than the denominator. Hence, Keq > 1.
11. (A) Right: 2 SO3 (g) + 2 Cl2 (g) <===> 2 SO2Cl2 (g) + O2 (g)
Number of moles on reactant side = 4
Number of moles on product side = 3
Change in number of moles = -1
For reactions accompanied by a decrease in number of moles from reactant side to product side, increasing the pressure will make the equilibrium shift to the right.
(B) Left: 2 NH3 (g) <===> N2 (g) + 3 H2 (g)
Change in number of moles from reactant to product side = +2
Increasing the pressure on such a system will shift the equilibrium to the left, i.e, more reactants will be produced.
(C) No effect: l2 (g) + H2 (g) <===> 2 Hl (g)
Change in number of moles from reactant side to product side = 0
Such systems are not affected by change in pressure.
(A) Right: O2 (g) + 2 F2 (g) <===> 2 OF2 (g)
Change in number of moles from reactant to product side = -1
Increasing pressure will shift the equilibrium to the right producing more products.
(C) No effect: CH4 (g) + 2 O2 (g) <===> CO2 (g) + 2 H2O (g)
Change in number of moles in going from reactant side to product side = 0
Increasing pressure will have no effect on the system at equilibrium.
(B) Left: 2 C8H18 (g) + 25 O2 <===> 16 CO2 (g) + 18 H2O (g)
Change in number of moles in going from the reactant to the product side = +7
Since the number of moles increases on the product side, increasing the pressure will shift the equilibrium position to the left, i.e., more reactants will be produced.
16. (B) False: Carboxylic acids and their derivatives like amides and carbonyl compounds can also form hydrogen bonds.
17. (A) True: Different kinds of enzymes are indeed needed to break down different kinds of substrates.
18. (A) True: Enzymes make a reaction proceed by lowering its activation energy which in turn depends on temperature.
19. (B) False: ROR is the condensed notation for ether. The condensed notation for ester is ROOR.
20. (A) True: These three structures give different information like bonding, helicity, etc.
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