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1 t is the same asA) Δ[A]/Δt B) -(5/4)(Δ[B]/Δt ) C) -(4/5)(Δ[B]/Δt ) D) -(1/4)(Δ[A]/Δt ) E) none of the above. 2 3 decomposes to a mixture of products. The following data were collected for the decomposition at 100º
C. What is the average rate of reaction, -Δ
[RX3 ]/Δt , over the entire experiment?<![CDATA[<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::::/sites/dl/free/0310402656/38225/ch16_1_02.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (0.0K)</a>]]> A) 0.0028 mol/L. s B) 0.045 mol/L. s C) 0.0014 mol/L. s D) -0.0027 mol/L. s E) 0.0057 mol/L. s 3 5 →
PCl3 + Cl2 is increased a factor of four when the concentration of PCl5 is doubled. Therefore, the rateA) depends on the concentrations of PCl3 and Cl2 . B) is first order with respect to PCl5 . C) is second order with respect to PCl5 . D) is fourth order with respect to PCl5 . E) is first order with respect to PCl3 . 4 iso -butane (iso -C4 H10 ) to form hydrogen bromide and the t -butyl radical (t -C4 H9 ): Br + iso -C4 H10 → HBr + t -C4 H9 . The reaction is first order with respect to each of the reactants, and at 298 K, the rate constant is 1.02 x 106 L mol-1 s-1 . Calculate the initial rate of the reaction in a 4.0 L reaction vessel containing 0.60 moles of iso -butane and 0.00010 moles of bromine atoms.A) 1.5 mol L-1 s-1 B) 3.8 mol L-1 s-1 C) 2.5 x 1011 mol L-1 s-1 D) 2.5 x 102 mol L-1 s-1 E) 6.7 x 102 mol L-1 s-1 5 4 H8 (g )) at 438º
C, C4 H8 (g ) → 2C2 H4 (g ).
The reaction follows first-order kinetics and the rate constant is k = 2.48 x 10-4 s-1 . If the initial concentration of cyclobutane is 0.800 mol/L, what concentration will remain after 10.0 min?A) 0.689 mol/L B) 0.455 mol/L C) 0.333 mol/L D) 0.248 mol/L E) 0.0061 mol/L 6 A) 9.1 x 10-3 min-1 B) 1.1 x 10-2 min-1 C) 31 min-1 D) 51 min-1 E) none of the above 7 A) ln[A] plotted against time gives a straight line of positive slope. B) ln[A] plotted against time gives a straight line of negative slope. C) 1/[A] plotted against time gives a straight line of positive slope. D) 1/[A] plotted against time gives a straight line of negative slope. E) None of the above plots is consistent with a second-order reaction. 8 A) [A] plotted against time gives a horizontal, straight line. B) ln[A] plotted against time gives a straight line of negative slope. C) 1/[A] plotted against time gives a straight line of positive slope. D) [A] plotted against time gives a straight line of negative slope. E) [A] plotted against time gives a curved line of negative slope, decreasing in magnitude as time increases. 9 A) 18.5 min-1 B) 38.5 min-1 C) 9.25 min-1 D) 19.3 min-1 E) 0.0260 min-1 10 -2 s-1 at 150°C, and the activation energy is 71 kJ/mol. What is the value of the rate constant at 170°C?A) 9.2 x 10-2 s-1 B) 3.7 x 10-2 s-1 C) 2.49 s-1 D) 4.0 x 10-2 s-1 E) none of the above 11 A) k ; T ; -E a B) k ; 1/T ; -E a C) lnk ; T ; -E a /R D) k ; 1/T ; -E a /R E) lnk ; 1/T ; -E a /R 12 A) Rate = k [2A] B) Rate = k [A] C) Rate = k [A]2 D) Rate = k [A]1/2 E) Rate = 2k [A] 13 - (aq ) → ClO3 - (aq )- (aq ) has been proposed to occur by the following mechanism.- (aq ) + ClO- ( aq )→ ClO2 - ( aq ) + Cl- ( aq ) (slow)2 - (aq ) + ClO- (aq ) → ClO3 - (aq ) + Cl- (aq ) (fast A) rate = k [ClO- ] B) rate = k [ClO- ]3 C) rate = k [ClO2 - ][ClO- ] D) rate = k [ClO- ]2 E) rate = k [Cl- ][ClO- ]2 14 g ) + 2H2 (g ) → N2 (g ) + 2H2 O(g )
g ) + NO(g ) → N2 O4 (g ) (slow)2 O4 (g ) + H2 (g ) → N2 (g ) + H2 O2 (g ) (fast)2 O2 (g ) + H2 (g ) → 2H2 O(g ) (fast)not affect the rate of the reaction?A) adding a catalyst B) adding more NO(g ) C) adding more H2 (g ) D) increasing the temperature E) All of these will affect the rate of the reaction. 15 A) increasing the number of high-energy molecules. B) increasing the temperature of the molecules in the reaction. C) increasing the number of collisions between molecules. D) increasing the activation energy for the reaction. E) providing a new reaction pathway for molecules.
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