NH3 1.26×10^4 g N2(g) + 3H2(g) → 2NH3(g) ΔH = –92.6 kJ

NH3 1.26×10^4 g N2(g) + 3H2(g) → 2NH3(g) ΔH = –92.6 kJ

 

 

암모니아 1.26×10^4 g 생성될 때 발생하는 열량은?

N2(g) + 3H2(g) → 2NH3(g) ... ΔH = –92.6 kJ/mol

 

 

Determine the amount of heat (in kJ) given off

when 1.26×10^4 g of ammonia is produced according to the equation:

N2(g) + 3H2(g) → 2NH3(g) ... ΔH = –92.6 kJ/mol

Assume the reaction takes place under standard-state conditions at 25℃.

 

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ΔH = –92.6 kJ은

2 mol NH3(g) 생성 시 반응열 이므로,

NH3 1 mol당 반응열을 계산하면,

–92.6 kJ / 2 mol = –46.3 kJ/mol

 

 

 

NH3의 몰질량 = 17 g/mol 이므로,

1.26×10^4 g / (17 g/mol) = 741.18 mol NH3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(–46.3 kJ/mol) × 741.18 mol = –34316.6 kJ

 

 

 

답: –3.43×10^4 kJ

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/17823 ]

1.26×10^4 g NO2 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH = –114.6 kJ