arrheniuspracticesolution_pdf

1 a 2 45 10 3 2 a 3 36 10 3 correct 3 a 1 79 10 3 4 a

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1. A = 2 . 45 × 10 3 2. A = 3 . 36 × 10 3 correct 3. A = 1 . 79 × 10 3 4. A = 5 . 71 × 10 2 5. A = 1 . 05 × 10 4 Explanation: k = A · e E a / ( R · T ) A = k e E a / ( R · T ) = 2 . 45 × 10 3 e 780 / (8 . 314 · 298) = 3 . 36 × 10 3 003 0.0points A given reaction has an activation energy of 24.52 kJ/mol. At 25 C the half-life is 4 min- utes. At what temperature will the half-life be reduced to 20 seconds? 1. 57.9 C 2. - 1 . 19 C 3. 25.5 C 4. 75.0 C 5. - 59 . 9 C 6. 125 C correct 7. 115 C 8. 100. C 9. 150 C Explanation: Use the Arrhenius equation. The 20 second reaction is running 12 times that of the 240 reaction. Put that ratio parenleftbigg 240 20 = 12 parenrightbigg in for
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casey (rmc2555) – non-graded arrhenius practice – holcombe – (51395) 2 the ln parenleftbigg k a k b parenrightbigg term. Remember to use Kelvin for the temperatures. 004 0.0points Consider the plot of ln k vs 1 T for a chemi- cal reaction. The activation energy for the reaction will be which of the following? 1. + slope of the plot R 2. + (slope of the plot) × R 3. - R slope of the plot 4. - slope of the plot R 5. - (slope of the plot) × R correct Explanation: The straight-line plot form of the Arrhenius equation is ln k = - E a R parenleftbigg 1 T parenrightbigg + ln A This matches the linear form of y = mx + b and the slope, m is equal to the - E a /R term.
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