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?

 

 

 

> 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

 

 

 

반응 S(s) + O2(g) → SO2(g)에서,

생성된 SO2 1 mol 당 ΔH = –226 kJ인 반응에 대해

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

b. 황 0.501 mol을 공기 중에서 태울 때 방출되는 열의 양을 계산하시오.

c. SO2(g) 1 mol을 구성 원소로 깨는 데 필요한 에너지의 양은 얼마인가?

답: a) –9.23 kJ; b) –148 kJ; c) +296 kJ/mol

 

 

For the reaction S(s) + O2(g) → SO2(g), ΔH = –226 kJ per mol of SO2 produced.

a. Calculate the amount of heat released when 1.00 g of sulfur is burned in the presence of oxygen.

b. Calculate the amount of heat released when 0.501 mol of sulfur is burned in air.

c. How much energy is needed to break 1 mol of SO2(g) into its constituent elements?

 

 

 

[FAQ] [①23/05/25] [②24/06/07]