When 1.00 L of 2.00 M strong acid (\"HA\") is mixed with 1.00 L of 2.00 M strong

Question

When 1.00 L of 2.00 M strong acid ("HA") is mixed with 1.00 L of 2.00 M strong base ("MOH") at 25.0oC, the temperature of the mixture increases to 27.7oC. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.184 J oC-1 g-1, and that the density of the final solution is 1.00 g/mL, calculate the enthalpy change per mole of water produced.

The following reaction of the "strong acid" and "strong base" is shown below:

HA + MOH --> MA + H2O

Explanation / Answer

volume of mixture of solutions = 1ml(strong acid )+1ml (strong base)= 2ml

density = 1 g/ml, mass of solution= volume* density= 2ml*1g/ml=2gm

temperature rise = 27.5-25= 2.5 deg.c

specific heat= 4.184 J/gm.degc

head added due to reaction= mass* specific heat* temperature difference= 2*4.184*2.5 =20.92 joules

since there is temperature rise, enthalpy change is -ve. Enthalpy change=-20.92 joules

moles iof HA= molarity* volume in Liters= 2*1/1000=0.002, moles of base= 2*1/1000 =0.002

the reaction is HA+ MOH -------->MA+ H2O

1 mole of HA reacts with 1 mole of MOH to give 1 mole of water

moles of water formed when 0.002 moles of HA and MOH are used =0.002

hence enthalpy change/mole=-20.92/0.002=10460 J/mole of water= 10.460 Kj/mole

since no heat is transferred to calorimeter, its change in enthalpy =0

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