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Darya [45]
3 years ago
13

I2 + Ti+4 ----> Ti + IO3- reducing agent

Chemistry
1 answer:
Schach [20]3 years ago
3 0

In order to identify the reducing agent let us check the oxidation states

I2 + Ti+4 ----> Ti + IO3- reducing agent

I2 : oxidation state of I = 0

Ti^+4 = oxidation state of Ti = +4

Ti : oxidation state = 0

IO3^-1 , oxidation state of Iodine = +5

So here the Ti undergoes reduction hence it is oxidizing agent

Where iodine undergoes oxidation hence it is reducing agent

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Answer:

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CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)

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Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}

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Using the information provided in lines 1 and 2 of the table:

0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1

Using the information provided in lines 3 and 4 of the table and insering the value of a:

0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\

0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5

The rate law is: Rate = k [CH_3 Cl] [Cl_2]^{0.5}

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