Question

Sodium hydroxide and HCl reat in a 1:1 ratio. If 20g of solid NaOH are added to 1000 mL of a solution containing .5 moles of HCl, the temp of the solution increases by 6C. Assuming the total solution mass is 1000g and the specific heat of the solution is 4.184, calculate the heat released by this reaction. Then caluclate delta H rxn (heat released per mole NaOH). Assume the density of solution is 1g/mL.

Answer 1

T = 6⁰C

specific of solution = 4.184 J/ g ⁰C

mass of solution = 1000 gm

q = mass of solution X specific heat of solution X T

= 1000 X 4.184 X 6

q = 25104 J = 25.104 KJ

heat released = 25.104 J

molar mass of NaOH = 39.997 gm/mol that mean 1 mole of NaOH = 39.997 gm

20 gm NaOH release heat = 25.104 KJ then 39.997 gm NaOH release heat = 39.997 X 25.104 / 20 = 50.204 KJ

heat is released in this reaction therefore have negative sign

H of reaction per mole of NaOH = - 50.204 KJ