You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a cof

Q: You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a cof

4. a. initial temperature(T1) = 22 C final temperature(T2) = 41.2 c DT = 41.2 - 22 = 19.2 c caloremeterconstant(C) = 109.2 j/c volume of reaction mixer = 50+50 = 100 ml mass of solution = d*V = 1*100 = 100 g specificheat of mixer = 4.184 j/g.c heat evolved(q) = m*s*DT + C*DT = 100*4.184*19.2 + 109*19.2 = 10126.08 joule = 10.126 kj b. molarheat of neutralisation = q/n = 10.126/(155*10^-3) = 65.33 kj/mol

ID: 1061934 • Letter: Y

Question

You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially a temperature of 22.0 degree C. The final temperature of the neutralization reaction was determined to be 41.2 degree C. The calorimeter constant was known to be 109.2 J/degree C. (Specific heat of H_2O = 4.184 J/g-degree C). What is the total amount of heat evolved in this reaction? If 155 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization for this reaction?

Explanation / Answer

a. initial temperature(T1) = 22 C

final temperature(T2) = 41.2 c

DT = 41.2 - 22 = 19.2 c

caloremeterconstant(C) = 109.2 j/c

volume of reaction mixer = 50+50 = 100 ml

mass of solution = d*V = 1*100 = 100 g

specificheat of mixer = 4.184 j/g.c

heat evolved(q) = m*s*DT + C*DT

= 100*4.184*19.2 + 109*19.2

= 10126.08 joule

= 10.126 kj

b. molarheat of neutralisation = q/n

= 10.126/(155*10^-3) = 65.33 kj/mol

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You mix 50.0 mL of a weak monoprotic acid wilh 50.0 ml. of NaOH solution in a co

You mix 50.0 ml of a weak monoprotic acid with 50.0 ml of NaOH solution in a cof

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You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a cof

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