1.07 M HCHO2 75.0 mL 22.4℃ 3.20 M NaOH 25.0 mL 23.4℃

1.07 M HCHO2 75.0 mL 22.4℃ 3.20 M NaOH 25.0 mL 23.4℃

 

 

반응열을 측정하기 위해

1.07 M HCHO2 75.0 mL를

22.4℃의 커피컵 열량계에 넣고

22.4℃의 3.20 M NaOH 25.0 mL를 가했다.

혼합물을 온도계로 빠르게 교반했더니,

온도가 23.4℃까지 상승하였다.

 

 

In the reaction between formic acid (HCHO2) and sodium hydroxide,

water and sodium formate (NaCHO2) are formed.

 

To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2

was placed in a coffee cup calorimeter at a temperature of 22.4℃,

and 25.0 mL of 3.20 M NaOH, also at 22.4℃, was added.

The mixture was stirred quickly with a thermometer,

and its temperature rose to 23.4℃.

 

a) Write the balanced chemical equation for the reaction.

b) Calculate the heat of reaction in joules.

Assume that the specific heats of all solutions are 4.18 J g-1 ℃-1

and that all densities are 1.00 g mL-1.

What is the heat of reaction per mole of acid (in units of kJ mol-1).

 

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HCHO2 + NaOH → NaCHO2 + H2O

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

= (4.18) (75.0 + 25.0) (23.4 – 22.4)

= 418 J

 

 

 

(1.07 mol/L) (75.0/1000 L) = 0.08025 mol HCHO2

 

 

 

418 J / 0.08025 mol = 5208.7 J/mol

= 5.21 kJ/mol

 

 

 

답: –5.21 kJ/mol

 

 

 

 

[참고] 산-염기 중화열 = 50 kJ/mol 이상

 

[ 관련 글 https://ywpop.tistory.com/3925 ] 산-염기 중화반응과 중화열

 

[ 관련 예제 https://ywpop.tistory.com/12917 ] 0.500 M HCl 200 mL + 0.500 M NaOH 200 mL 중화열 계산

 

 

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1.07 M HCHO2 75.0 mL 20.8℃ 1.78 M NaOH 45.0 mL 25.3℃

 

To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2

was placed in a coffee cup calorimeter at a temperature of 20.8 °C,

and 45.0 mL of 1.78 M NaOH, also at 20.8 °C, was added.

The mixture was stirred quickly with a thermometer,

and its temperature rose to 25.3 °C.

 

 

q = C m Δt

= (4.18) (75.0 + 45.0) (25.3 – 20.8)

= 2257.2 J

 

 

(1.07 mol/L) (75.0/1000 L) = 0.08025 mol HCHO2

 

 

2257.2 J / 0.08025 mol = 28127 J/mol

= 28.1 kJ/mol