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…

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.

MARKS Content Team · Accepted Answer

Correct answer: (C) -1652.8Cal. logK_2K_1 = (∆H)/2.303R(1T_1-1T_2Where K_2 is equilibrium constant at higher temperature = 0.60 ×10^(-4)K_1 is equilibrium constant at lower temperature = 1.2 ×10^(-4)T_2 = 127°C = 400KT_1 = 27 °C = 300 K∆H = Heat of formation of ammonia per mole = ?log0.61.2 = ∆H2.303×1.98 cal1)/300-1/400∆H = (2.303×1.98×1200×(-0.30))/1 = -1652.8 cal

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.

Solution

Relevant Questions

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.

Solution

Relevant Questions

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.