0.500 M H2SO4 25.0 mL 1.00 M KOH 25.0 mL 23.50℃ 30.17℃

0.500 M H2SO4 25.0 mL 1.00 M KOH 25.0 mL 23.50℃ 30.17℃

 

 

When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50℃, the temperature rises to 30.17℃.

Calculate ΔH of this reaction.

Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water (1.00 g/mL and 4.184 J/g•K).

 

 

생성된 물 1 mol당 ΔH를 kJ 단위로 구하라.

 

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H^+ 이온의 몰수를 계산하면,

2 × (0.500 mol/L) × (25.0/1000 L) = 0.0250 mol H^+

( H2SO4는 H^+ 이온을 2개 내놓는 2가산 )

H2SO4(aq) → 2H^+(aq) + SO4^2-(aq)

 

 

OH^- 이온의 몰수를 계산하면,

1 × (1.00 mol/L) × (25.0/1000 L) = 0.0250 mol OH^-

KOH(aq) → K^+(aq) + OH^-(aq)

 

 

 

0.0250 mol H^+ = 0.0250 mol OH^-

이므로, 용액은 완전 중화됨.

H^+(aq) + OH^-(aq) → H2O(l)

---> 0.0250 mol H2O 생성.

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

= (4.184) (25.0 + 25.0) (30.17 – 23.50)

= 1395 J

---> ΔH = -1395 J

 

 

 

0.0250 mol H2O 생성되므로,

H2O 1몰당 엔탈피를 계산하면,

ΔH = -1395 J / 0.0250 mol

= -55800 J/mol

= -55.8 kJ/mol

 

 

 

답: -55.8 kJ/mol