Thermochemistry -enthalpy change of neutralisation. HCL (aq) + NaOH (aq)---> NaCL(aq) = H2O?

Question

250 cm3 of a solution of NaOH of concentration 0.400 moldm -3 was placed in a polystyrene cup and the temp recorded to the nearest 0.1 oC.

250 cm3 of a solution of HCL of concentration 0.400 moldm-3 was placed in another polystyrene cup and the temp of the solution recorded to the nearest 0.1oC. The temp of both solutions was found to be 17.4oC.

On mixing the two solutions the temp of rose to a max of 20oC. Assuming the specific heat capacity of all the solutions is 4.2 JoC-1 G-1. Complete the steps to calculate the enthalpy change of neutralisation.

-MASS= g

-RISE IN TEMP OF SOLUTION= 2.6 oC (20-17.4 =2.6)

-ENERGY GAINED BY THE SOLUTION Q=

-Number of moles of water formed=

-Standard enthalpy of neutralisation=

Any help with this would be greatly appreciated!

Thanks

Answer 1

a. 500

b. 2.6

c. 500 x 4.2 x 2.6 (and then make it negative)

d. (250/1000) x 0.4

e. divide (c) by (d).  Then divide by 1000 to get kJ/mol.  Should be around -57.