A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO
ID: 1009547 • Letter: A
Question
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO_3 in a coffee cup calorimeter. If both solutions were initially at 35.00degreeC and the temperature of the resulting solution was recorded as 37.00degreeC, determine the delta H degree_rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water. (Use 1.0 g/mL as the density of the solution and 4.2 J/(gK) as the specific heat capacity of the solution)Explanation / Answer
No of mol of NaOH = 0.1*0.3 = 0.03 mol
No of mol of HNO3 = 0.1*0.3 = 0.03 mol
q = msDT
m = mass of mixer = 200*1 = 200 grams
s = specific heat = 4.18 j/g.c
Dt = 37-35 = 2 c
q = 200*4.18*2
= 1672 j
= 1.672 kj
DHrxn = -1.672 kj
DHrxn per mol = -1.672/0.03 = -55.73 kj/mol
as both NaOH,HCl is strong like NAoH,HNO3.so that
DHrxn = -55.73 kj/mol
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.