Kg mol 500 518 1 344988 o C this is the CHANGE T bp T new T normal T new T bp T

Kg mol 500 518 1 344988 o c this is the change t bp t

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Kg mol 500 . 0 518 . 0 (1) = 3.44988 o C this is the CHANGE T bp = T new T normal T new = T bp + T normal = 110.6 + 3.44988 o C 114 o C ans (d) 1
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6. answer (e) is correct . A rate law cannot be determined by inspection of the overall balance equation for the bulk chemical reaction. Answers (a), (b), (c), and (d) all treat the overall reaction equation like a mechanism step, and try to take the rate law relationships from the stoichiometry. This is not appropriate because the mechanism is not provided. 7. Use the integrated rate law for a 1 st order reaction, ln kt amt starting left amt - = All of the variables are provided except amt left. Substitute the known values and solve. Ln = 0314 . 0 left amt (6.4 x 10 3 min 1 )(62.0 min) It is helpful to recall that Ln = 0314 . 0 left amt ln (amt left) ln (0.0314) ln (amt left) ( 3.46095) = 0.3968 amt left = 0.02112 M ans. (d) 8. Let “cat” mean “catalyst” and “no cat” means “no catalyst” eqn (i) ln k with cat = ln A ( 29 cat with a E RT , 1 eqn (ii) ln k no cat = ln A ( 29 cat no a E RT , 1 Subtract (ii) from (i): ln k with cat ln k no cat = 0 ( 29 cat with a E RT , 1 + ( 29 cat no a E RT , 1 recall that ln k with cat ln k no cat = ln cat no cat with k k which is the ratio requested. ln cat no cat with k k = ( 29 kJ kJ K K mol kJ x 55 64 ) 673 )( / 10 3145 . 8 ( 1 3 - - ln (ratio) = 1.698 ratio = e 1.698 = 4.9947 5.0 ans (a) 9. The plot of 1/[X] vs. time is linear with positive slope. This is diagnostic for 2 nd order kinetics. If rate is only dependent on [X] then the term [Y] does not need to appear in the rate law but if it did, it would have an exponent of 0. Only answer (b) is correct . 10. (a) wrong. Increasing temperature on a reaction not at equilibrium has no effect on the concentration of reactants. (Had equilibrium been established, and had Temp THEN been increased, for an exothermic reaction more reactants would form over time. But that is not this question. This question is about kinetics.) (b) Wrong. Increasing Temp. means that more molecules will have the threshold E a for reaction, but the Ea itself is a constant that depends on the molecular structure, which includesmlowest energy configuration, and probable alternative configurations. Molecular structure does not change with temperature. (c) Correct Collision frequency increases because the average particle velocity is greater at higher temps, they travel faster and run into each other more often. (d) Correct . The fraction of collisions with sufficient energy increases as more particles have energy > Ea 2
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(e) wrong. See answer (b) (f) wrong. It is true that the rate of internal rotations may increase, but the statistical probability that a given molecule has a given orientation at any point in time remains the same. 11. Perform algebraic manipulation on the equations (i) and (ii) until their sum yields equation (iii). Keep eqn. (i) as is, and its Keq as is. Keep eqn. (ii) as is; add the two equations and multiply their K’s, to find K overall .
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