ID: C 16, The half-life of the reaction A B is 0.560 minutes, and the rate-law e
ID: 102672 • Letter: I
Question
ID: C 16, The half-life of the reaction A B is 0.560 minutes, and the rate-law expression has the form: rate-k If the initial concentration of A is 3.40 M, what is the rate constant? a. 3.04 mo/L min b. 6.07 mol/L·min c. 0.357 mol/L·min d. 1.79 mol/L min e. 1.25 mol/L .min Molten AcCl, is electrolyzed for 5.0 hours with 17. at one electrode and chlorine gas, Cl, is produced at the other. How many grams of actinium are produced? a. 5.6g b. 0.22 g e. 0.67 g d. 4.4g e. 1.1 g Which one of the following combinations cannot be a buffer solution? 18. b. HNO,-NaNO, c. NH,- (NH) SO d. NH,- NH NO, e. HCIO -NaCIO Rate data have been determined at a particular temperature for the overall reaction 2NO + 2H2 N2 + 2H20 in which all reactants and products are gases. 19. Trial Run Initial [NO] Initial [H.] Initial Rate (Msl) 0.10M 020 00150 0.10 M 0.30 M 0.20 M 0.0225 0.0600 0.20 M a. rate HNOJ[H:] b. rate = kfNOF[H12 c. rate kINOF[H.] d. rate KNOJ[H-P e. None of the preceding answers is correct 20. For a particular reaction at 25"C, ar--197 kImol, and AS - -113,3 J'mol-k. At which of the following would the reaction be spontaneous? a. 3750 K b. 1050 K c. 2750 K d. 3250 K e. 2450 kExplanation / Answer
16. with rate = k
k = rate constant
this is a zero-order reaction
For zero-order reaction, half-life t1/2 is,
t1/2 = [Ao]/2k
with,
[Ao] = 3.40 M
t1/2 = 0.560 min
we get,
rate constant k = 3.40/2 x 0.560 = 3.04 mol/L.min
a. 3.04 mol/L.min
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17. Charge Q = It
I = 5 h x 60 min x 60 s
t = 0.31 A
So,
Q = 5 x 60 x 60 x 0.32 = 5580 C
moles electron used = 5580/96500 = 0.058 mol e-
1 mole of electron would deposit 1/3 mole of Ac
so,
moles of Ac deposited = 0.058/3 = 0.0193 mol
mass of Ac deposited = 0.0193 mol x 227 g/mol = 4.4 g
d. 4.4 g
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18. The combination which cannot be a buffer is,
b. HNO3-NaNO3
A buffer is a combination of weak acid-conjugate base ot weak base-conjugate acid. In b. we have strong acid and thus this cannot form buffer.
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19. From 1 and 2, [NO] = constant
when concentration of [H2] was increased by 1.5 times, the rate also increased by 1.5 times,
so order with respect to [H2] = 1
From 1 and 3, [H2] = constant
when concentration of [NO] was doubled, the rate quadrupled
So the order with respect to [NO] = 2
The rate expression thus becomes,
c. rate = k[NO]^2.[H2]
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20. with dHo = -197 kJ/mol and dSo = -113.3 J/K.mol
with dGo = 0
dGo = dHo - TdSo
Temperature (T) = dHo/dSo = -197/-0.1133 = 1739 K
So below 1739 K the reaction would have lower -TdSo value, so dGo would be -ve and spontaneous reaction.
b. 1050 K