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# Answer to Question #268422 in Chemical Engineering for Tj2

Question #268422

A reaction between methanol and oxygen can be used to produce formaldehyde: 2πΆπ»3ππ» + π2 (π) β 2π»πΆπ»π(π) + 2π»2π(π): βπ»Μπ π = β326.2 ππ½/πππ The standard heat of combustion of hydrogen is: π»2 (π) + 1 2 π2 (π) β π»2π(π): βπ»Μπ π = β285.8 ππ½/πππ (a) Use these heats of reaction and Hessβs law to determine the standard heat of the direct decomposition of methanol to form formaldehyde: πΆπ»3ππ»(π) β π»πΆπ»π(π) + π»2(π)

1
2021-11-19T00:07:11-0500

There are total pf 3 reactions:

(1) 2CH3OH + O2 --> 2HCHO + 2H2O "(\\Delta{H}^0_r=-326.2\\ kJ\/mol)"

(2) H2 + 1/2O2 --> H2O "(\\Delta{H}^0_r=-285.8\\ kJ\/mol)"

(3) CH3OH --> HCHO + H2

To obtain reaction (3) as an algebraic sum of reactions (1) and (2) according to Hess's law, the reaction (1) should be divided by 2, while reaction (2) should be reversed. The same conversions apply to corresponding heats of each reaction: the heat of reaction (1) is divided by 2, and heat of reaction (2) changes its sign. As a result we get:

(1) CH3OH + 1/2O2 --> HCHO + H2O "(\\Delta{H}^0_r=-163.1\\ kJ\/mol)"

(2) H2O --> H2 + 1/2O2 "(\\Delta{H}^0_r=285.8\\ kJ\/mol)"

The desired reaction (3) is now an algebraic sum of reactions (1) and (2):

CH3OH + Β ΜΆ1ΜΆ/ΜΆ2ΜΆOΜΆ2ΜΆ + Β ΜΆHΜΆ2ΜΆOΜΆ --> HCHO + Β ΜΆHΜΆ2ΜΆOΜΆ + H2 + Β ΜΆ1ΜΆ/ΜΆ2ΜΆOΜΆ2ΜΆ

Therefore,

"\\Delta{H^0_r(3)}=\\Delta{H^0_r(1)}+\\Delta{H^0_r(2)}=-163.1+285.8=122.7\\ kJ\/mol"

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