The equilibrium constant Kp for the formation of ammonia from its constituent elements at 27°C is 1.2×10-4 and at 127°C is 0.60×10-4. Calculate the mean heat of formation of ammonia per mole in this temperature range.

TS EAMCET · Chemistry · Chemical Equilibrium

The equilibrium constant $(K_{p})$ for the formation of ammonia from its constituent elements at $27 ° C$ is $1.2 \times 10^{- 4}$ and at $127 ° C$ is $0.60 \times 10^{- 4} .$ Calculate the mean heat of formation of ammonia per mole in this temperature range.

A $- 82.64 Cal$
B $- 826.4 Cal$
C $- 1652.8 Cal$
D $- 165.2 Cal$

Verified Solution

Answer & Solution

Correct Answer

(C) $- 1652.8 Cal$

Step-by-step Solution

Detailed explanation

logK2K1 = ∆H2.303R1T1-1T2 Where K2 is equilibrium constant at higher temperature = 0.60 ×10-4K1 is equilibrium constant at lower temperature = 1.2 ×10-4T2 =…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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