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Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)
Correct answer is between '-20.90,-18.20'. Can you explain this answer?
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Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chlorof...
To calculate the free energy change of the reaction, we can use the equation:
ΔG = -RT ln(Kc)
Where:
ΔG = free energy change of the reaction (in Latm)
R = ideal gas constant (0.082 LatmK⁻¹mol⁻¹)
T = temperature in Kelvin (291K)
ln = natural logarithm
Kc = equilibrium constant (1.14)
First, let's convert the concentration from molarity to partial pressure. In an ideal gas, the partial pressure is directly proportional to the concentration. Since the concentration of both gases is 0.5 mol/L, we can assume that the partial pressure of each gas is also 0.5 atm.
Now, let's substitute the values into the equation:
ΔG = - (0.082 LatmK⁻¹mol⁻¹) * (291K) * ln(1.14)
Using a calculator, we can evaluate the natural logarithm of 1.14 and multiply it by -0.082 * 291 to find the free energy change of the reaction.
ΔG = - (0.082 LatmK⁻¹mol⁻¹) * (291K) * ln(1.14)
≈ -35.60 Latm
ΔG = -RT ln(Kc)
Where:
ΔG = free energy change of the reaction (in Latm)
R = ideal gas constant (0.082 LatmK⁻¹mol⁻¹)
T = temperature in Kelvin (291K)
ln = natural logarithm
Kc = equilibrium constant (1.14)
First, let's convert the concentration from molarity to partial pressure. In an ideal gas, the partial pressure is directly proportional to the concentration. Since the concentration of both gases is 0.5 mol/L, we can assume that the partial pressure of each gas is also 0.5 atm.
Now, let's substitute the values into the equation:
ΔG = - (0.082 LatmK⁻¹mol⁻¹) * (291K) * ln(1.14)
Using a calculator, we can evaluate the natural logarithm of 1.14 and multiply it by -0.082 * 291 to find the free energy change of the reaction.
ΔG = - (0.082 LatmK⁻¹mol⁻¹) * (291K) * ln(1.14)
≈ -35.60 Latm
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Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer?
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Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer?.
Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer?.
Solutions for Kc for the equilibrium reaction N2O4 dissociating into 2NO2 in chloroform at 291K is 1.14. Calculate free energy change of reaction (Latm) when concentration of two gases are 0.5 mol/L each at same temp. (R = 0.082 LatmK–1mol–1, up to two decimal places)Correct answer is between '-20.90,-18.20'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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