The equilibrium constant (K_p ) 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…

Question

The equilibrium constant (K_p ) 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.

Accepted Answer

Correct answer: B. (b) log ⁡ K_2K_1= - H2.303R( 1T_1- 1T_2) Substituting the values in above equation log ⁡ 0.6 10^-41.2 10^-4= - H2.303 2( 1300- 1400) ⇒ log ⁡0.5=- H4.606 ( 100120000 ) ⇒ H=-1663.69 Mean heat of formation = H2= -1663.692 ≈-826.43 cal

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

Solution

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