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Parents Pricing Home SSCE VCE Chemistry Carbon-Based Fuels Coke, which is essentially pure carbon, is widely used as a fuel
Coke, which is essentially pure carbon, is widely used as a fuel - VCE - SSCE Chemistry - Question 3 - 2005 - Paper 1 Question 3
View full question Coke, which is essentially pure carbon, is widely used as a fuel. Its complete combustion can be represented by the following equation.
C(s) + O2(g) → CO2(g) ΔH = ... show full transcript
View marking scheme Worked Solution & Example Answer:Coke, which is essentially pure carbon, is widely used as a fuel - VCE - SSCE Chemistry - Question 3 - 2005 - Paper 1
Calculate the energy released when 80% of coke is oxidised to carbon dioxide Only available for registered users.
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First, convert the mass of coke to moles:
e x t m a s s o f c o k e = 2.00 e x t t o n n e = 2.00 i m e s 1 0 6 e x t g r a m s ext{mass of coke} = 2.00 ext{ tonne} = 2.00 imes 10^6 ext{ grams} e x t ma sso f co k e = 2.00 e x t t o nn e = 2.00 im es 1 0 6 e x t g r am s n = rac{2.00 imes 10^6 ext{ grams}}{12 ext{ g/mol}} = 1.67 imes 10^5 ext{ moles}
Since 80% is oxidised to CO₂:
e x t m o l e s o f C O 2 = 0.80 i m e s 1.67 i m e s 1 0 5 = 1.34 i m e s 1 0 5 e x t m o l e s ext{moles of CO}_2 = 0.80 imes 1.67 imes 10^5 = 1.34 imes 10^5 ext{ moles} e x t m o l eso f CO 2 = 0.80 im es 1.67 im es 1 0 5 = 1.34 im es 1 0 5 e x t m o l es
Using the enthalpy of formation for CO₂:
e x t E n e r g y = 1.34 i m e s 1 0 5 e x t m o l e s i m e s − 393 e x t k J / m o l = − 5.27 i m e s 1 0 7 e x t k J ext{Energy} = 1.34 imes 10^5 ext{ moles} imes -393 ext{ kJ/mol} = -5.27 imes 10^7 ext{ kJ} e x t E n er g y = 1.34 im es 1 0 5 e x t m o l es im es − 393 e x t k J / m o l = − 5.27 im es 1 0 7 e x t k J
Calculate the energy released when 20% of coke is oxidised to carbon monoxide Only available for registered users.
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Next, calculate the moles of coke oxidised to CO:
20% of coke:
e x t m o l e s o f C O = 0.20 i m e s 1.67 i m e s 1 0 5 = 3.34 i m e s 1 0 4 e x t m o l e s ext{moles of CO} = 0.20 imes 1.67 imes 10^5 = 3.34 imes 10^4 ext{ moles} e x t m o l eso f CO = 0.20 im es 1.67 im es 1 0 5 = 3.34 im es 1 0 4 e x t m o l es
Using the enthalpy of formation for CO:
e x t E n e r g y = 3.34 i m e s 1 0 4 e x t m o l e s i m e s − 232 e x t k J / m o l = − 7.75 i m e s 1 0 6 e x t k J ext{Energy} = 3.34 imes 10^4 ext{ moles} imes -232 ext{ kJ/mol} = -7.75 imes 10^6 ext{ kJ} e x t E n er g y = 3.34 im es 1 0 4 e x t m o l es im es − 232 e x t k J / m o l = − 7.75 im es 1 0 6 e x t k J
Calculate total energy released Only available for registered users.
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Finally, sum the energies:
e x t T o t a l E n e r g y = − 5.27 i m e s 1 0 7 e x t k J + − 7.75 i m e s 1 0 6 e x t k J = − 6.05 i m e s 1 0 7 e x t k J ext{Total Energy} = -5.27 imes 10^7 ext{ kJ} + -7.75 imes 10^6 ext{ kJ} = -6.05 imes 10^7 ext{ kJ} e x t T o t a lE n er g y = − 5.27 im es 1 0 7 e x t k J + − 7.75 im es 1 0 6 e x t k J = − 6.05 im es 1 0 7 e x t k J
In conclusion, the energy released is approximately 60.5 million kJ.
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