EXPERIMENT.

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Presentation transcript:

EXPERIMENT

RESULTS TIME (SECONDS) TEMPERATURE (OC) 20 5 35 10 55 15 50 MAXIMUM TEMPERATURE Question 1 - Calculate the energy released in your reaction. The density of water is 1 g/cm3 . The specific heat capacity of water is 4.2 J/g/°C. ANSWER DH = mcDT = 25 X 4.2 X (55 – 20) = 3675J = 3.675kJ

ANSWER The Zinc was in excess QUESTION 2 - Calculate the moles of zinc that you used. ANSWER REMEMBER From C2 (Additional Science) MOLES = MASS ÷ Mr ∴ MOLES = 2 ÷ 65 = 0.0308 mols of Zinc QUESTION 3 - The moles of CuSO4 solution that was used was 0.0125mols. Which substance Zn or CuSO4 was in excess? ANSWER The Zinc was in excess

ANSWER For Zinc = 3.675 ÷ 0.0308 = 119kJ/mol QUESTION 4 - Calculate the energy change in kJ/mol for the chemicals reacting ANSWER For Zinc = 3.675 ÷ 0.0308 = 119kJ/mol For CuSO4 = 3.675 ÷ 0.0125 = 294kJ/mol

a) Calculate the energy released in this reaction in kJ QUESTION -5g of zinc was added to 75cm3 of 1moldm-3 solution of CuSO4 solution. The temperature rose from 25 to 670C. a) Calculate the energy released in this reaction in kJ b) Calculate the moles of Zinc metals used c) The moles of CuSO4 used in this experiment was 0.075mols. Which substance Zinc or CuSO4 was in excess? d) Calculate the energy change in kJ/mol for the chemicals reacting ANSWER DH = mcDT = 75 x 4.2 x (67 – 25) = = 13230J = 13.23kJ b) Moles = Mass ÷ Mr = 5/65 = 0.0769 mols Zinc c)The zinc was in excess d) For Zinc = 13.23 ÷ 65 = 0.204kJ/mol For CuSO4 = 13.23 ÷ 0.075 = 176.4kJ/mol

REMEMBER METAL HYDROXIDE + ACID  SALT + WATER This is a neutralisation reaction

Question 1- Calculate the energy released in your reaction Question 1- Calculate the energy released in your reaction. The density of water is 1 g/cm3 . The specific heat capacity of water is 4.2 J/g/°C. ANSWER DH = m x c x DT = 50 x 4.2 x 28 = 5880J = 5.88kJ Question 2 -The number of moles of the acid used was 0.025mols. Calculate the energy released in the reaction in kJ/mol ANSWER 5.88/0.025 = 235kJ/mol

The energy released in the reaction in kJ QUESTION - 75cm3 of 0.5moldm-3 HCl reacts with 75cm3 of 0.5moldm-3 NaOH. The temperature rose from 20 to 420C. Calculate: The energy released in the reaction in kJ The moles of HCl used in this reaction is 0.0375mols. Calculate the energy released in the reaction in kJ/mol ANSWERS 13.86kJ 369.6kJ/mol

A BOMB CALORIMETER