50.0 mL 0.400 M Ca(NO3)2 50.0 mL 0.800 M NaF CaF2 25.00℃

50.0 mL 0.400 M Ca(NO3)2 50.0 mL 0.800 M NaF CaF2 25.00℃

 

 

When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF,

CaF2 precipitates, as shown in the net ionic equation below.

The initial temperature of both solutions is 25.00℃.

Assuming that the reaction goes to completion,

and that the resulting solution has a mass of 100.00 g

and a specific heat of 4.18 J/(g•℃),

calculate the final temperature of the solution.

Ca^2+(aq) + 2F^-(aq) → CaF2(s) ... ΔH° = –11.5 kJ

 

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각 반응물의 몰수를 계산하면,

> (0.400 mol/L) (50.0 mL) = 20 mmol Ca(NO3)2

> (0.800 mol/L) (50.0 mL) = 40 mmol NaF

 

 

 

Ca^2+(aq) + 2F^-(aq) → CaF2(s)

 

Ca^2+ : F^- = 1 : 2 계수비(= 몰수비) 이므로,

Ca^2+ : F^- = 1 : 2 = 20 mmol : ? mmol

? = 2 × 20 = 40 mmol F^-

---> Ca(NO3)2와 NaF는 전부 반응.

 

 

 

Ca^2+ : CaF2 = 1 : 1 계수비(= 몰수비) 이므로,

생성된 CaF2의 몰수 = 20 mmol

 

 

 

Ca^2+(aq) + 2F^-(aq) → CaF2(s) ... ΔH° = –11.5 kJ

---> CaF2 1 mol당 11.5 kJ의 열을 방출하므로,

 

20 mmol CaF2에 의해 방출되는 열은

(11500 J/mol) (20/1000 mol) = 230 J

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

Δt = q / C m

= 230 J / [(4.18 J/(g•℃)) (100.00 g)]

= 0.55℃

 

 

 

25.00 + 0.55 = 25.55℃

 

 

 

답: 25.55℃