1]Consider the following balanced reaction:
\(2A_2 + 3B \rightarrow C_2 + 2D \)
What is the enthalpy of reaction if the enthalpy of formation of the compounds is given as
\(A_2 = 7.78 kJ/mol; B = -18.43 kJ/mol; C_2 = 0 kJ/mol; D = 9.96 kJ/mol.\)
Record your answer in scientific notation using 3 significant figures.
My answer to this question is 59.65 kJ/mol computed as follows: [9.96*2-2*(7.78)-3*(-18.43)=59.65]
2]Calculate ΔH° for the process Co3O4(s) → 3 Co(s) + 2 O2(g)
from the following information
Co(s) + ½ O2(g) → CoO(s) ΔH° = -120.6 kJ/mol
3 CoO(s) + ½ O2(g) → Co3O4(s) ΔH° = -632.4 kJ/mol
I am working on these two questions. Any chemistry help will be accepted.
Thermochemistry question part (2)
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- Dhamnekar Winod
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Thermochemistry question part (2)
Last edited by Dhamnekar Winod on Thu Jul 01, 2021 3:31 am, edited 1 time in total.
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Re: Thermochemistry question part (2)
The first one is -59.65 kJ/ mol educt- product.
The second one 3*(-120,6 kJ/mol) + (-632,4 kJ/mol) = -994,2 kJ/ mol
The second one 3*(-120,6 kJ/mol) + (-632,4 kJ/mol) = -994,2 kJ/ mol
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Re: Thermochemistry question part (2)
Both the reactions are endothermic , but your negative signs implies both reactions are exothermic. How is that?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Thermochemistry question part (2)
Who said that. In first it was told your result is wrong.
So take the equation half, the Addition of educt is one enthalpy value. The product also have value, what has subtracted vom educt value, it gives negative value.
The cobaltoxide reaction.
First CoO is formed with negativ value, from there Co3O4 is formed also with negativ value. The summery is still negativ. But it is the decomposition asked, that it is then the opposit positive.
So take the equation half, the Addition of educt is one enthalpy value. The product also have value, what has subtracted vom educt value, it gives negative value.
The cobaltoxide reaction.
First CoO is formed with negativ value, from there Co3O4 is formed also with negativ value. The summery is still negativ. But it is the decomposition asked, that it is then the opposit positive.
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Re: Thermochemistry question part (2)
Can I ask an off-topic question? I want to buy an infrared thermometer but don't know a good brand, can you advise me? Thanks.