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헤스법칙. C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(l) 반응엔탈피 계산

by 영원파란 2020. 4. 28.

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헤스법칙. C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2H2O(l) 반응엔탈피 계산

 

 

The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water.

C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2H2O(l)

 

Calculate ΔH for this reaction from the following data.

C6H4(OH)2(aq) C6H4O2(aq) + H2(g) ... ΔH° = +177 kJ/mol

H2O(l) + 1/2 O2(g) H2O2(aq) ... ΔH° = +94.6 kJ/mol

H2(g) + 1/2 O2(g) H2O(l) ... ΔH° = -286 kJ/mol

 

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참고: 헤스의 법칙 [ https://ywpop.tistory.com/3376 ]

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C6H4(OH)2(aq) C6H4O2(aq) + H2(g) ... ΔH° = +177 kJ/mol

H2O2(aq) H2O(l) + 1/2 O2(g) ... ΔH° = -94.6 kJ/mol

H2(g) + 1/2 O2(g) H2O(l) ... ΔH° = -286 kJ/mol

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C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2H2O(l) ... ΔH° = ?

 

 

ΔH° = (+177) + (-94.6) + (-286) = -203.6 kJ/mol

 

 

: -203.6 kJ/mol

 

 

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