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Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1 Question 6
View full question Consider the following information.
• Ethanol burns in excess air according to the following equation.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔH = -1364 kJ mol^-1
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View marking scheme Worked Solution & Example Answer:Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1
Calculate the minimum amount of energy, in kJ, required to heat 550 g of water and the pot from 18.5°C to 100.0°C. Only available for registered users.
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To calculate the total energy required, we must sum the energy needed to heat both the water and the aluminum pot.
Energy for Water:
Using the formula:
Q = m c r i a n g l e T Q = mc riangle T Q = m cr ian g l e T
where:
m m m = mass of water = 550 g,
c c c = specific heat capacity of water = 4.18 J g^{-1} °C^{-1},
r i a n g l e T riangle T r ian g l e T = change in temperature = 100.0°C - 18.5°C = 81.5°C.
Q w a t e r = 550 i m e s 4.18 i m e s 81.5 = 198 , 165 e x t J = 198.2 e x t k J Q_{water} = 550 imes 4.18 imes 81.5 = 198,165 ext{ J} = 198.2 ext{ kJ} Q w a t er = 550 im es 4.18 im es 81.5 = 198 , 165 e x t J = 198.2 e x t k J
Energy for Aluminium Pot:
For the aluminum pot, we use the same formula:
where:
m m m = mass of aluminum = 150 g,
c c c = specific heat capacity of aluminum = 9.000 J g^{-1} °C^{-1}.
Q a l u m i n i u m = 150 i m e s 9.000 i m e s 81.5 = 1 , 116 , 750 e x t J = 1116.8 e x t k J Q_{aluminium} = 150 imes 9.000 imes 81.5 = 1,116,750 ext{ J} = 1116.8 ext{ kJ} Q a l u mini u m = 150 im es 9.000 im es 81.5 = 1 , 116 , 750 e x t J = 1116.8 e x t k J
Total Energy Required:
Q t o t a l = Q w a t e r + Q a l u m i n i u m = 198.2 + 1116.8 = 1315 e x t k J Q_{total} = Q_{water} + Q_{aluminium} = 198.2 + 1116.8 = 1315 ext{ kJ} Q t o t a l = Q w a t er + Q a l u mini u m = 198.2 + 1116.8 = 1315 e x t k J
Thus, the minimum energy required is approximately 1315 kJ .
Calculate the mass, in g, of ethanol that needs to be completely burnt to provide this energy. Only available for registered users.
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To determine the mass of ethanol required to provide this energy:
Use the heat of combustion of ethanol:
Given that the heat of combustion for ethanol is ΔH = -1364 kJ/mol.
Calculate the moles of ethanol needed:
Using the formula:
ext{moles} = rac{Q_{needed}}{ ext{ΔH}} = rac{1315 ext{ kJ}}{1364 ext{ kJ/mol}} = 0.965 ext{ mol}
Calculate the mass of ethanol:
The molar mass of ethanol (C2H5OH) is 46.07 g/mol.
e x t m a s s = e x t m o l e s i m e s e x t m o l a r m a s s = 0.965 i m e s 46.07 = 44.42 e x t g ext{mass} = ext{moles} imes ext{molar mass} = 0.965 imes 46.07 = 44.42 ext{ g} e x t ma ss = e x t m o l es im ese x t m o l a r ma ss = 0.965 im es 46.07 = 44.42 e x t g
Thus, 44.42 g of ethanol needs to be burnt.
Calculate the mass, in g, of ethanol that needs to be burnt in practice to heat the water and the pot from 18.5°C to 100.0°C. Only available for registered users.
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In this part, we take into account that only 35% of the energy released from the combustion of ethanol is effectively used:
Calculate the effective energy released:
Q e f f e c t i v e = 0.35 i m e s Q n e e d e d = 0.35 i m e s 1315 e x t k J = 460.25 e x t k J Q_{effective} = 0.35 imes Q_{needed} = 0.35 imes 1315 ext{ kJ} = 460.25 ext{ kJ} Q e ff ec t i v e = 0.35 im es Q n ee d e d = 0.35 im es 1315 e x t k J = 460.25 e x t k J
Calculate the moles of ethanol for effective energy:
ext{moles} = rac{Q_{effective}}{ ext{ΔH}} = rac{460.25 ext{ kJ}}{1364 ext{ kJ/mol}} = 0.337 ext{ mol}
Calculate the mass of ethanol burned in practice:
e x t m a s s = e x t m o l e s i m e s e x t m o l a r m a s s = 0.337 i m e s 46.07 = 15.53 e x t g ext{mass} = ext{moles} imes ext{molar mass} = 0.337 imes 46.07 = 15.53 ext{ g} e x t ma ss = e x t m o l es im ese x t m o l a r ma ss = 0.337 im es 46.07 = 15.53 e x t g
Thus, 15.53 g of ethanol needs to be burnt in practice.
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