0.844 M HCl 200 mL 0.422 M Ba(OH)2 200 mL 27.5℃

0.844 M HCl 200 mL 0.422 M Ba(OH)2 200 mL 27.5℃

 

 

200 mL of a 0.844 M HCl aqueous solution is mixed with

200 mL of 0.422 M Ba(OH)2 aqueous solution

in a coffee-cup calorimeter.

Both the solutions have an initial temperature of 27.5 °C.

Calculate the final temperature of the resulting solution,

given the following information:

H^+(aq) + OH^-(aq) → H2O(l) ... ΔH_rxn = –56.2 kJ/mol

Assume that volumes can be added,

that the density of the solution

is the same as that of water (1.00 g/mL),

and the specific heat of the solution

is the same as that for pure water, 4.184 J/g•K.

 

---------------------------------------------------

 

> (0.844 mol/L) (0.200 L) = 0.1688 mol HCl

( 참고 https://ywpop.tistory.com/7787 )

 

= 0.1328 mol H^+

 

 

 

> 0.422 × 0.200 = 0.0844 mol Ba(OH)2

> 2 × 0.0844 = 0.1688 mol OH^-

= H^+의 몰수

 

 

 

(–56.2 kJ/mol) (0.1688 mol) = –9.48656 kJ

---> (–)부호는 방출열, 즉 발열 반응을 의미.

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

Δt = q / C m

= (9486.56 J) / [(4.184 J/g•℃) (400 g)]

= 5.67℃

 

 

 

27.5 + 5.67 = 33.2℃

 

 

 

답: 33.2℃

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/22786 ]

0.664 M HCl 200 mL 0.332 M Ba(OH)2 200 mL 31.2℃

 

 

 

[키워드] HCl + Ba(OH)2 반응 기준문서