헤스의 법칙. 2B(s) + 3H2(g) → B2H6(g)

헤스의 법칙. 2B(s) + 3H2(g) → B2H6(g)

 

 

Given the following data:

a) 4B(s) + 3O2(g) → 2B2O3(s) ... ΔH = -2509.1 kJ

b) 2H2(g) + O2(g) → 2H2O(l) ... ΔH = -571.7 kJ

c) B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l) ... ΔH = -2147.5 kJ

 

Calculate ΔH for the reaction: 2B(s) + 3H2(g) → B2H6(g)

 

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▶ 참고: 헤스의 법칙 https://ywpop.tistory.com/3376

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a’) 2B(s) + 3/2 O2(g) → B2O3(s) ... ΔH = (-2509.1)/2 kJ

b’) 2(3/2) H2(g) + 3/2 O2(g) → 2(3/2) H2O(l) ... ΔH = (-571.7)(3/2) kJ

c’) B2O3(s) + 3H2O(l) → B2H6(g) + 3O2(g) ... ΔH = +2147.5 kJ

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2B(s) + 3H2(g) → B2H6(g)

 

 

((-2509.1)/2) + ((-571.7)(3/2)) + (+2147.5) = +35.4 kJ

 

 

답: +35.4 kJ

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/14943 ] ΔH 계산. 2B(s) + 3H2(g) → B2H6(g)