S(s) + O2(g) → SO2(g) ΔH = –296 kJ/mol

S(s) + O2(g) → SO2(g) ΔH = –296 kJ/mol

1.00 g S 0.501 mol S 1 mol SO2(g)

 

 

For the reaction S(s) + O2(g) → SO2(g),

ΔH = –296 kJ/mol of SO2 formed.

 

a) Calculate the quantity of heat released

when 1.00 g of sulfur is burned in oxygen.

 

b) Calculate the quantity of heat released

when 0.501 mole of sulfur is burned in air.

 

c) What quantity of energy is required

to break up 1 mole of SO2(g) into its constituent elements?

 

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> S의 몰질량 = 32.0650 ≒ 32.07 g/mol

 

 

 

a) Calculate the quantity of heat released

when 1.00 g of sulfur is burned in oxygen.

 

1.00 g / (32.07 g/mol) = 0.03118 mol S

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

ΔH = 0.03118 mol × (–296 kJ/mol) = –9.23 kJ

 

 

 

 

b) Calculate the quantity of heat released

when 0.501 mole of sulfur is burned in air.

 

ΔH = 0.501 mol × (–296 kJ/mol) = –148 kJ

 

 

 

 

c) What quantity of energy is required

to break up 1 mole of SO2(g) into its constituent elements?

 

S(s) + O2(g) → SO2(g) ... ΔH = –296 kJ/mol

의 역반응이므로,

 

SO2(g) → S(s) + O2(g) ... ΔH = +296 kJ/mol

 

 

 

ΔH = 1 mol × (+296 kJ/mol) = +296 kJ

 

 

 

 

[키워드] 번역 오타 기준문서, 번역 오타 문제 기준문서, S(s) + O2(g) → SO2(g) 1 mol ΔH = –226 kJ S 1.00 g S 0.501 mol

 

S(s) + O2(g) → SO2(g) ... ΔH = –226 kJ/mol

황 1.00 g을 산소 존재하에서 태울 때 방출되는 열의 양을 계산하시오.

 

(1.00 g) (1 mol / 32.07 g) (–226 kJ/mol) = –7.05 kJ

 

 

 

[FAQ] [①23/05/25]