When 50.0 ml of 1.00 M HCL was mixed with 50.0 mL of 1.00 M NaOH in a Calorimete

Question

When 50.0 ml of 1.00 M HCL was mixed with 50.0 mL of 1.00 M NaOH in a Calorimeter, the temperature rose from 25 degree Celsius to 31 degree Celsius. Calculate the heat for the reaction, assuming that the total volume of the solution is 100.0 ml of density 1.0 g/ml and specifi heat 1.00 cal/( g. degree Celsius) a) Write a balance equation for the reaction b) How many moles of HCL (aq) and NaOH have reacted c) Is the reaction exothermic or endothermic d) How many calories are evolved in the reaction e) What is the change H for the above reaction I KJ/ mol

Explanation / Answer

a) Hcl(aq) + NaOH(aq) --------------> NaCl (aq) + H2O(l)

b) Hcl(aq) + NaOH(aq) --------------> NaCl (aq) + H2O(l)

50x1.00 50x1.00

Thus moles of Hcl = moles of NaOh reacted = 50x1.00/1000 = 0.005 moles

= 5 x10-3 moles

c) the reaction is exothermic as the temperature of calorimeter is increased from 25 to 31 C

neutralisation is an exothermic reaction.

d) the heat released in the reaction = mass of solution x specific heat x rise in temperature

= 100g x 1.00cal/g.degree x 6 degree

= 600 cal.

e) delta H of the reaction

0.005 moles of acid ad base on neutralisation gave 600 cal of heat

1 moles of acid and base on neutralisation can give = 1 mol x 600cal /0.005 mol

= 120000cal

= 120 kcal/mol

As 1 cal = 4.18 J

the heat of reaction = 120x 4.18kJ

= 501.6kJ/mol

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