tión enthalpies Consider the following unbalanced reaction: ? Al(s) +? Fe203 (s) ===> Al,03 (s) + 2 Fe(s), AHrxn=? Thermodynamic data at 298.15 K: AHf(Fe203) = -822.16 kļ/mol, AHf(Al203) = -1669.8 k/mol. %3D The specific heat capacity of copper (Cu)-0.376812 g K), assumed to be constant. The molar mass of aluminum and Al203 is 26.98153849 and 101.961276 g/mol, respectively. (1) Calculate AHrxn from the formation enthalpies. Is the rn exothermic or endothermic? Solution: -847.64 kJ. Exothermic. AHrxn = (-1669.8)-(-822.16)=-847.64 kJ per 2 mol of Al (2) Calculate the reaction enthalpy per gram of Al(s). Solution: -15.71 k/ gram of AI. (-(847.64 / 2) kJ/mol of Al)/ (26.98 g/mol of Al)-. -15.71 k/gram of Al. (3) Assume that only 20 % of the released energy can be used to heat a metal (the 80 % remaining is wasted energy). How many grams of Al (s) are needed to heat a copper (Cu) mass of 1 kg by 100 Celsius? Solution: 12 g of Al(s).
tión enthalpies Consider the following unbalanced reaction: ? Al(s) +? Fe203 (s) ===> Al,03 (s) + 2 Fe(s), AHrxn=? Thermodynamic data at 298.15 K: AHf(Fe203) = -822.16 kļ/mol, AHf(Al203) = -1669.8 k/mol. %3D The specific heat capacity of copper (Cu)-0.376812 g K), assumed to be constant. The molar mass of aluminum and Al203 is 26.98153849 and 101.961276 g/mol, respectively. (1) Calculate AHrxn from the formation enthalpies. Is the rn exothermic or endothermic? Solution: -847.64 kJ. Exothermic. AHrxn = (-1669.8)-(-822.16)=-847.64 kJ per 2 mol of Al (2) Calculate the reaction enthalpy per gram of Al(s). Solution: -15.71 k/ gram of AI. (-(847.64 / 2) kJ/mol of Al)/ (26.98 g/mol of Al)-. -15.71 k/gram of Al. (3) Assume that only 20 % of the released energy can be used to heat a metal (the 80 % remaining is wasted energy). How many grams of Al (s) are needed to heat a copper (Cu) mass of 1 kg by 100 Celsius? Solution: 12 g of Al(s).
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![problem on reaction enthalpies
Consider the following unbalanced reaction:
? Al(s) + ? Fe203 (s) ===> Al203 (s) + 2 Fe(s), AHrxn=?
Thermodynamic data at 298.15 K: AHf(Fe203) = -822.16 kj/mol, AH(Al203) = -1669.8 kJ/mol.
%3D
The specific heat capacity of copper (Cu)=0.376812 Jg K), assumed to be constant. The molar mass of aluminum and Al203 is 26.98153849 and
101.961276 g/mol, respectively.
(1) Calculate AHrxn from the formation enthalpies. Is the rxn exothermic or endothermic? Solution: -847.64
kJ. Exothermic.
AHrxn = (-1669.8)-(-822.16)%=-847.64 kJ per 2 mol of Al
(2) Calculate the reaction enthalpy per gram of Al(s). Solution: -15.71 kj / gram of Al.
(-(847.64 / 2) kJ/mol of Al) / (26.98 g/mol of Al) =-15.71 kigram of Al.
(3) Assume that only 20 % of the released energy can be used to heat a metal (the 80 % remaining is wasted energy). How many grams of Al (s) are
needed to heat a copper (Cu) mass of 1 kg by 100 Celsius? Solution: 12 g of Al(s).
(4) How many grams of Al 20 3 (s) are produced in part (3) ? Solution: 22.66 g of Al203.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44a73c06-f6ec-4388-a28b-a0d09afcf92c%2F45215463-17c8-4760-a1f2-fcb810c7c056%2F992v58o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:problem on reaction enthalpies
Consider the following unbalanced reaction:
? Al(s) + ? Fe203 (s) ===> Al203 (s) + 2 Fe(s), AHrxn=?
Thermodynamic data at 298.15 K: AHf(Fe203) = -822.16 kj/mol, AH(Al203) = -1669.8 kJ/mol.
%3D
The specific heat capacity of copper (Cu)=0.376812 Jg K), assumed to be constant. The molar mass of aluminum and Al203 is 26.98153849 and
101.961276 g/mol, respectively.
(1) Calculate AHrxn from the formation enthalpies. Is the rxn exothermic or endothermic? Solution: -847.64
kJ. Exothermic.
AHrxn = (-1669.8)-(-822.16)%=-847.64 kJ per 2 mol of Al
(2) Calculate the reaction enthalpy per gram of Al(s). Solution: -15.71 kj / gram of Al.
(-(847.64 / 2) kJ/mol of Al) / (26.98 g/mol of Al) =-15.71 kigram of Al.
(3) Assume that only 20 % of the released energy can be used to heat a metal (the 80 % remaining is wasted energy). How many grams of Al (s) are
needed to heat a copper (Cu) mass of 1 kg by 100 Celsius? Solution: 12 g of Al(s).
(4) How many grams of Al 20 3 (s) are produced in part (3) ? Solution: 22.66 g of Al203.
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