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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?
a)1250k
b)500 k
c)1000k
d)750k?
a)1250k
b)500 k
c)1000k
d)750k?
Most Upvoted Answer
Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectivel...
Calculating the Equilibrium Temperature
Given:
Standard entropy of X2 = 60 J/K/mol
Standard entropy of Y2 = 40 J/K/mol
Standard entropy of XY3 = 50 J/K/mol
Enthalpy change for the reaction 1/2 X2 + 3/2Y2 → XY3 = -30 kJ/mol
To calculate the equilibrium temperature of the reaction, we can use the following formula:
ΔG° = -RT ln(K)
Where:
ΔG° = Standard free energy change
R = Gas constant (8.314 J/K/mol)
T = Temperature in Kelvin
ln(K) = Natural logarithm of the equilibrium constant (K)
At equilibrium, ΔG° = 0, and we can rearrange the formula to solve for T:
T = -ΔG° / (R ln(K))
Calculating the Equilibrium Constant
The equilibrium constant (K) can be calculated using the following formula:
ΔG° = -RT ln(K)
K = e^(-ΔG° / (RT))
To calculate ΔG°, we need to use the following formula:
ΔG° = ΔH° - TΔS°
Where:
ΔH° = Enthalpy change
ΔS° = Standard entropy change
Substituting the given values, we get:
ΔH° = -30 kJ/mol
ΔS° = (1/2)(60 J/K/mol) + (3/2)(40 J/K/mol) - (1)(50 J/K/mol) = 40 J/K/mol
ΔG° = -30,000 J/mol - T(40 J/K/mol)
At equilibrium, ΔG° = 0, so:
0 = -30,000 J/mol - T(40 J/K/mol)
T = -30,000 J/mol / (40 J/K/mol)
T = 750 K
Therefore, the equilibrium temperature of the reaction is 750 K, which is option (d).
Given:
Standard entropy of X2 = 60 J/K/mol
Standard entropy of Y2 = 40 J/K/mol
Standard entropy of XY3 = 50 J/K/mol
Enthalpy change for the reaction 1/2 X2 + 3/2Y2 → XY3 = -30 kJ/mol
To calculate the equilibrium temperature of the reaction, we can use the following formula:
ΔG° = -RT ln(K)
Where:
ΔG° = Standard free energy change
R = Gas constant (8.314 J/K/mol)
T = Temperature in Kelvin
ln(K) = Natural logarithm of the equilibrium constant (K)
At equilibrium, ΔG° = 0, and we can rearrange the formula to solve for T:
T = -ΔG° / (R ln(K))
Calculating the Equilibrium Constant
The equilibrium constant (K) can be calculated using the following formula:
ΔG° = -RT ln(K)
K = e^(-ΔG° / (RT))
To calculate ΔG°, we need to use the following formula:
ΔG° = ΔH° - TΔS°
Where:
ΔH° = Enthalpy change
ΔS° = Standard entropy change
Substituting the given values, we get:
ΔH° = -30 kJ/mol
ΔS° = (1/2)(60 J/K/mol) + (3/2)(40 J/K/mol) - (1)(50 J/K/mol) = 40 J/K/mol
ΔG° = -30,000 J/mol - T(40 J/K/mol)
At equilibrium, ΔG° = 0, so:
0 = -30,000 J/mol - T(40 J/K/mol)
T = -30,000 J/mol / (40 J/K/mol)
T = 750 K
Therefore, the equilibrium temperature of the reaction is 750 K, which is option (d).
Community Answer
Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectivel...
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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k?
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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k?.
Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k?.
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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k?, a detailed solution for Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? has been provided alongside types of Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k? theory, EduRev gives you an
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