Question

Enter your numbers into 1 decimal places. For the reaction:

4NO(g) + 6H₂O(g) ⇋ 4NH3(g) + 5O₂(g)

1. Determine ΔH° and ΔS° for the reaction at 298 K.

   $LaTeX: \Delta$ ΔH⁰(kJ)

   $LaTeX: \Delta$ ΔS⁰(J/K)

2. Assuming that these values are relatively independent of temperature, calculate ΔG° at 100.°C and 2560.°C.

   $LaTeX: \Delta$ ΔG⁰(kJ)at 100℃  

$LaTeX: \Delta$ ΔG⁰(kJ) at 2560℃

3. Is this reaction spontaneous at all range of temperature? Enter 1 for true or 0 for false.

  

4. Calculate the K at 298K for the above reaction.  Enter "is" for infinitely small and "ig" infinitely great.

K at 298K

Answer 1

To find out enthalpy and entropy change of the reaction we need enthalpy of formation and standard entropy of all the products and reactant at 298 K respectively.

Change in enthalpy of reaction = Net enthalpy of formation of products - Net enthalpy of formation of reactant

Change in standard entropy = Net standard entropy of products - Net standard entropy of reactant

We know that , DG = DH - TDS

Given assumption is that change in enthalpy and standard entropy of the reaction is independent of temperature.

We have, for equilibrium reaction

DG = - RT lnK

- 4 NOL) + 6H₂O (8) At 298 K - 4 NH₃(g) +50219 AS (J/K mol) 4/H" (kJ/mol Substance 90.29 -241.818 -45.90 NO (9) H₂O (9) NH₂ (8) 0,01) 210.8 188.8 193.8 +88.8 205.2 Ax H = 44, Hº (NH, (n) +5 4; H*(0,19) - 44, 4°(NO (1) - 60,#* (8,00) AH : 898-148 kJ at 298 K - Ho = 906.2 KJ AS = -174.8 J/K 06 = Auto - TAS A60 (373 k) = 963.3 kJ AGO (2833 k) = 1401.4 kJ AGO (298 k) = 958.3 kJ Ink AGO = -RT LnK 7 958.3 = - 8.314 x 105 x 298 - Ink = -386.8 & K = 1 x 10-168 Hence , K is infinitely small at 298k.

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