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Given the mechanism for the thermal decomposition of acetaldehyde, $C{ H }_{ 3 }CHO$:

$$C{ H }_{ 3 }CHO \overset { { k }_{ 1 } }{ \rightarrow } C{ H }_{ 3 }\cdotp + CHO\cdotp$$

$$C{ H }_{ 3 }\cdotp + C{ H }_{ 3 }CHO \overset { { k }_{ 2 } }{ \rightarrow } C{ H }_{ 4 } + C{ H }_{ 2 }CHO\cdotp$$

$$C{ H }_{ 2 }CHO\cdotp \overset { { k }_{ 3} }{ \rightarrow } CO + C{ H }_{ 3 }\cdotp$$

$$C{ H }_{ 3 }\cdotp + C{ H }_{ 3 }\cdotp \overset { { k }_{ 4 } }{ \rightarrow } { C }_{ 2 }{ H }_{ 6 }$$

Using the steady-state approximation for the intermediates, $C{ H }_{ 3 }\cdotp$ and $C{ H }_{ 2 }CHO\cdotp$, which of the following gives the rate law for methane ($C{ H }_{ 4 }$) formation?

A

$\frac { d[C{ H }_{ 4 }] }{ dt } = { k }_{ 2 }[C{ H }_{ 3 }\cdotp][C{ H }_{ 3 }CHO]$

B

$\frac { d[C{ H }_{ 4 }] }{ dt } ={ k }_{ 1 }{ [C{ H }_{ 3 }CHO] }$

C

$\frac { d[C{ H }_{ 4 }] }{ dt } = { k }_{ 2 }{ (\frac { { k }_{ 1 } }{ {2k }_{ 4 } } ) }^{ 1/2 }{ [C{ H }_{ 3 }CHO] }^{ 1/2 }$

D

$\frac { d[C{ H }_{ 4 }] }{ dt } = { k }_{ 2 }{ (\frac { { k }_{ 1 } }{ { 2k }_{ 4 } } ) }^{ 1/2 }{ [C{ H }_{ 3 }CHO] }^{ 3/2 }$

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