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
°
𝐶4 𝐻10(𝑔) +
𝑂
→ 4 𝐶𝑂2(𝑔) + 5 𝐻2 𝑂(𝑙)
∆𝐻𝑐𝑜𝑚𝑏
= −2877 𝑘𝐽/𝑚𝑜𝑙
2 2(𝑔)
First, we need to know the amount of heat generated from 1.50 g of butane:
°
𝑞𝑟 = ∆𝐻 = 𝑛∆𝐻𝑐𝑜𝑚𝑏
=
°
(1.50 𝑔)(−2877000 𝐽𝑚𝑜𝑙 −1 )
𝑚∆𝐻𝑐𝑜𝑚𝑏
=
= −74249 𝐽
(4 × 12.0107𝑔𝑚𝑜𝑙 −1 + 10 × 1.00794𝑔𝑚𝑜𝑙 −1 )
𝑀
Next, we can use this heat to determine the heat gained by the surroundings, which is the
water and the beaker:
𝑞𝑠𝑦𝑠 = −𝑞𝑠𝑢𝑟 = −(𝑞𝑤 + 𝑞𝑏 ) = −[𝑚𝑤 𝑐𝑤 (𝑇𝑓 − 𝑇𝑖 ) + 𝑐𝑏 (𝑇𝑓 − 𝑇𝑖 )]
−𝑞𝑠𝑦𝑠 = (𝑚𝑤 𝑐𝑤 + 𝑐𝑏 )(𝑇𝑓 − 𝑇𝑖 )
𝑇𝑖 =
𝑇𝑖 =
𝑞𝑠𝑦𝑠
+ 𝑇𝑓
(𝑚𝑤 𝑐𝑤 + 𝑐𝑏 )
(−74249 𝐽)
+ 70.5℃ = 24.568℃ = 24.6℃
[(350.0𝑔)(4.19𝐽𝑔−1 ℃−1 ) + (150.0 𝐽℃−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.
𝐶4 𝐻10(𝑔) +
13
𝑂
→ 4 𝐶𝑂2(𝑔) + 5 𝐻2 𝑂(𝑙)
2 2(𝑔)
P.E. (kJ)
𝐶4 𝐻10(𝑔) ,
°
∆𝐻𝑐𝑜𝑚𝑏
= −2877 𝑘𝐽/𝑚𝑜𝑙
13
𝑂
2 2(𝑔)
°
∆𝐻𝑐𝑜𝑚𝑏
= −2877 𝑘𝐽
4 𝐶𝑂2(𝑔) , 5 𝐻2 𝑂(𝑙)
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).