61.1 mL 0.543 M HCl(aq) 42.6 mL NaOH(aq) 17.8℃ 21.6℃

61.1 mL 0.543 M HCl(aq) 42.6 mL NaOH(aq) 17.8℃ 21.6℃

 

 

A chemist wants to determine the enthalpy of neutralization

for the following reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

She uses a calorimeter to neutralize completely

61.1 mL of 0.543 mol/L HCl(aq) with 42.6 mL of NaOH(aq).

The initial temperature of both solution is 17.8°C,

and the final temperature is 21.6°C.

Calculate the enthalpy of neutralization in units of kJ/mol of HCl.

Assume the density of both solutions is 1.00 g/mL,

and that the specific heat capacity of both solutions

is the same as that of water.

 

 

The chemist uses a simple calorimeter to neutralize completely

61.1 mL of 0.543 mol/L HCl(aq) with

42.6 mL of sufficiently concentrated NaOH(aq).

The initial temperature of both solutions is 17.80 °C.

After neutralization, the highest recorded temperature is 21.60 °C.

Calculate the enthalpy of neutralization in kJ/mol of HCl(aq).

 

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▶ 참고: 산-염기 중화반응과 중화열

[ https://ywpop.tistory.com/3925 ]

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sufficiently concentrated NaOH(aq)

---> HCl이 한계 반응물.

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

 

 

m = 61.1 + 42.6 = 103.7 g

 

Δt = 21.6 – 17.8 = 3.8℃

 

 

 

q = (4.184 J/g•℃) (103.7 g) (3.8 ℃)

= 1648.747 J

= 1.648747 kJ

 

 

 

(0.543 mol/L) (61.1/1000 L)

= 0.033177 mol HCl

 

 

 

1.648747 kJ / 0.033177 mol

= 49.7 kJ/mol

---> HCl 1몰당 중화열

 

 

 

답: 49.7 kJ/mol

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/22281 ]

1 M HCl 50 mL + 2 M NaOH 50 mL. 20℃ to 26℃

 

 

 

[키워드] HCl + NaOH 중화열 기준문서, HCl + NaOH 중화열 계산 기준문서