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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...
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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.
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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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