SCH4U0 Calorimetry and Enthalpy Practice Quiz 1) a) If 1.50 g of butane gas is used to heat 350.0 mL of water in a glass beaker to 70.5°C, what was the initial temperature of the water? The beaker has a heat capacity of 150.0 J/°C. 13 ° πΆ4 π»10(π) + π → 4 πΆπ2(π) + 5 π»2 π(π) βπ»ππππ = −2877 ππ½/πππ 2 2(π) First, we need to know the amount of heat generated from 1.50 g of butane: ° ππ = βπ» = πβπ»ππππ = ° (1.50 π)(−2877000 π½πππ −1 ) πβπ»ππππ = = −74249 π½ (4 × 12.0107ππππ −1 + 10 × 1.00794ππππ −1 ) π Next, we can use this heat to determine the heat gained by the surroundings, which is the water and the beaker: ππ π¦π = −ππ π’π = −(ππ€ + ππ ) = −[ππ€ ππ€ (ππ − ππ ) + ππ (ππ − ππ )] −ππ π¦π = (ππ€ ππ€ + ππ )(ππ − ππ ) ππ = ππ = ππ π¦π + ππ (ππ€ ππ€ + ππ ) (−74249 π½) + 70.5β = 24.568β = 24.6β [(350.0π)(4.19π½π−1 β−1 ) + (150.0 π½β−1 )] ∴ the initial temperature of both the water and beaker was 24.6ºC. 1) b) Using the same reaction from a) draw a potential energy diagram to illustrate the enthalpy change. Label the products, reactants, and enthalpy difference. πΆ4 π»10(π) + 13 π → 4 πΆπ2(π) + 5 π»2 π(π) 2 2(π) P.E. (kJ) πΆ4 π»10(π) , ° βπ»ππππ = −2877 ππ½/πππ 13 π 2 2(π) ° βπ»ππππ = −2877 ππ½ 4 πΆπ2(π) , 5 π»2 π(π) R.C. c) Explain what type of system (closed, open, or isolated) is described in a). Give reasons for your choice. The system is open because beakers do not have caps. The water in the beaker is exposed to the air, allowing both heat and matter to flow into and out of the beaker. d) Explain why there is a change in enthalpy when this reaction takes place. Why is there a change in enthalpy during a state change? There is a change in enthalpy because the reactants and products do not have the same potential energies. For example, the bonds are not the same, they involve different numbers of protons/electrons and different bond distances which will result in different amounts of energy being released when they are formed. For this reason a different quantity of energy is required to break the bonds in the reactants than is released when the bonds in the products are formed. Also, since the reactants and products will have different polarities, the strength of their intermolecular forces will also be different. This results in different amounts of energy being released/absorbed when the reactant/product solutes interact with the solvent (or themselves, or other solutes).