Using Hess’s Law and your results from parts I and II, calculate ÄH rxn, for rea

Question

Using Hess’s Law and your results from parts I and II, calculate ÄH rxn, for reaction CH3COOH(aq) ->CH3COO (aq) + H (aq) results:

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PlEASEEEE someone answer this correctly and explain. I actually want to learn, I dont pay like $15 a month for this site just to get answers only. I want an answer, how to get the answer and a clear explanation pls

Explanation / Answer

NaOH + HCl -----> NaCl + H2O

NaOH + CH3COOH -----> CH3COONa + H2O

we can rewrite this equations as

OH- + H+ ----> H2O ------> 1st

OH- + CH3COOH -----> CH3COO- + H2O -------> 2nd

2nd + reverse reaction of 1st =====>

CH3COOH ----> H+ + CH3COO-

It implies enthalpy of 2nd reaction + enthalpy of reverse reaction of first reaction gives the resultant enthalpy

Enthalpy of reaction = m*C*dT

m = mass of solution

C = heat capacity = 4.18

dT = change in temeprature

1st reaction enthalpy = 99.42*4.18*5 = 2077.878 J

It is exothermic reaction ; so delta H1 = -2077.878 J

reverse reaction of 1st reaction is ---->

delta H = 2077.878 J

for 2nd reaction ---->

delta H2 = m*C*dT = 99.684*4.184*4 = 1668.3 J

it is also exothermic reaction

so delta H2 = -1668.3 J

Important note ---> Acid + Base reaction is ----> exothermic reaction

Net enthalpy for the reaction = -1668.3 + 2077.878 = 409.566 J

Answer is +409.566 J

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