Use the References to access important values if needed for this question The fo
ID: 1027638 • Letter: U
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Use the References to access important values if needed for this question The following initial rate data are for the reaction of tertiary butyl bromide with hydroxide ion at 55 °C: (CH3)3CBr+OH(CH COH +Br Experiment 0.366 0.366 0.732 732 0.256 512 0.256 0.512 Initial Rate, Ms 4.06× 10-3 4.06x103 8.13x103 13x10-3 Complete the rate law for this reaction in the box below Use the form kIAIIBI, where 'I' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n Rate From these data, the rate constant is Submit Answer Retry Entire Group 5 more group attempts remainingExplanation / Answer
For the given reaction,
(CH3)3CBr + OH- ---> (CH3)3COH + Br-
studying the data in the table,
From Experiment 1 and 3, [OH-] = constant, [(CH3)3CBr] is doubled, the rate also doubled, so the order with respect to [(CH3)3CBr] = 1
From Experiment 1 and 2, [OH-] = doubled, [(CH3)3CBr] = constant, the rate remained the same, so the order with respect to [OH-] = 0
The rate eqation thus would be,
rate = k[(CH3)3CBr]
Rate constant k = rate/[(CH3)3CBr]
= 4.06 x 10^-3 M.s-1/0.366 M
= 0.0111 s-1
----
For the oxidation of arsenate ion by Ce(IV),
studying the data in the table,
From Experiment 1 and 2, [Ce^4+] = constant, [AsO3^3-] is doubled, the rate also doubled, so the order with respect to [AsO3^3-] = 1
From Experiment 1 and 3, [Ce^4+] = doubled, [AsO3^3-] = constant, the rate increased four times, so the order with respect to [Ce^4+] = 2
The rate eqation thus would be,
rate = k[Ce^4+][AsO3^3-]^2
Rate constant k = rate/[Ce^4+][AsO3^3-]^2
= 2.93 x 10^-3 M.s-1/(1.62 x 10^-2 M)(0.569 M)^2
= 0.559 M^-2.s^-1
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