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What will be the equilibrium constant at 127degree C .if equilibrium constant at 27 degree C is 4 for reaction N2 3H2<---->2NH3 ∆H=-46.06kj?
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What will be the equilibrium constant at 127degree C .if equilibrium c...
Introduction:
The given reaction is N2 + 3H2 ↔ 2NH3, and its equilibrium constant at 27°C is 4. The enthalpy change (∆H) for the reaction is -46.06 kJ. We need to find the equilibrium constant at 127°C.
Explanation:
To find the equilibrium constant at 127°C, we can use the Van't Hoff equation, which relates the equilibrium constant at different temperatures to the enthalpy change of the reaction.
ln (K2/K1) = ∆H/R [(1/T1) - (1/T2)]
where K1 is the equilibrium constant at temperature T1, K2 is the equilibrium constant at temperature T2, ∆H is the enthalpy change of the reaction, R is the gas constant (8.314 J/mol K), and T1 and T2 are the temperatures in Kelvin.
We can rearrange the equation to solve for K2:
K2 = K1 exp[∆H/R ((1/T1) - (1/T2))]
Plugging in the given values, we get:
K2 = 4 exp[-46060/(8.314*300) ((1/300) - (1/400))]
K2 = 7.05
Therefore, the equilibrium constant at 127°C is 7.05.
Conclusion:
The equilibrium constant at 127°C for the given reaction is 7.05. We used the Van't Hoff equation to relate the equilibrium constant at different temperatures to the enthalpy change of the reaction.
The given reaction is N2 + 3H2 ↔ 2NH3, and its equilibrium constant at 27°C is 4. The enthalpy change (∆H) for the reaction is -46.06 kJ. We need to find the equilibrium constant at 127°C.
Explanation:
To find the equilibrium constant at 127°C, we can use the Van't Hoff equation, which relates the equilibrium constant at different temperatures to the enthalpy change of the reaction.
ln (K2/K1) = ∆H/R [(1/T1) - (1/T2)]
where K1 is the equilibrium constant at temperature T1, K2 is the equilibrium constant at temperature T2, ∆H is the enthalpy change of the reaction, R is the gas constant (8.314 J/mol K), and T1 and T2 are the temperatures in Kelvin.
We can rearrange the equation to solve for K2:
K2 = K1 exp[∆H/R ((1/T1) - (1/T2))]
Plugging in the given values, we get:
K2 = 4 exp[-46060/(8.314*300) ((1/300) - (1/400))]
K2 = 7.05
Therefore, the equilibrium constant at 127°C is 7.05.
Conclusion:
The equilibrium constant at 127°C for the given reaction is 7.05. We used the Van't Hoff equation to relate the equilibrium constant at different temperatures to the enthalpy change of the reaction.
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What will be the equilibrium constant at 127degree C .if equilibrium constant at 27 degree C is 4 for reaction N2 3H22NH3 ∆H=-46.06kj?
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