From the calculations performed, the free energy change for the reaction is 72 kJ/mol.
<h3>What is the equilibrium constant?</h3>
The equilibrium constant is a value that shows the extent to which reactants have been converted to products.
Given that the equation of the reaction is;
3CH4(g)→C3H8(g)+2H2(g)
Then;
PC3H8 = 0.013 atm
PH2 = 2.3×10−2 atm
PCH4 = 41 atm
Now;
ΔG = ΔG° + RTlnQ
ΔG°reaction = ΔG°products - ΔG°reactants
ΔG°reaction = [( -23.4) +2(0)] - 3(-50.8)
ΔG°reaction = 129 kJ/mol
Q = PC3H8 * PH2^2/PCH4^3
Q = 0.013 * (2.3×10−2)^2/( 41)^3
Q = 6.877 * 10^-6/68921
Q= 9.9* 10^-11
Hence;
ΔG = 129 * 10^3 + [8.314 * 298 * (ln 9.9* 10^-11 )]
ΔG = 129 * 10^3 - 57073
ΔG = 72 kJ/mol
Learn more about free energy change: brainly.com/question/14143095
A
and it is D for your ansower
Explanation:
2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.
Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.
A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.
Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.
Full question :
Q. Which reactant is unlikely to produce the indicated product upon strong heating?
- A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
- B) 2-Ethylpropanedioic acid Butanoic acid
- C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
- D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
- E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone
Hence, option (D) is correct.
Learn more about carboxylic acid here : brainly.com/question/26855500
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Answer:
Final temperature = 83.1 °C
Explanation:
Given data:
Mass of concrete = 25 g
Specific heat capacity = 0.210 cal/g. °C
Initial temperature = 25°C
Calories gain = 305 cal
Final temperature = ?
Solution:
Q = m. c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C
305 cal = 5.25cal/°C × T2 - 25°C
305 cal / 5.25cal/°C = T2 - 25°C
58.1 °C = T2 - 25°C
T2 = 58.1 °C + 25°C
T2 = 83.1 °C
