Part 1: Using the table in the introduction, calculate the value of ?H in units

Question

Part 1: Using the table in the introduction, calculate the value of ?H in units of kJ/mol. After, calculate the value of ?S in units of J/(mol*K). Finally, cCalculate the value of ?G in units of kJ/mol for the reaction at 298 K (watch your units).

Part 2: Which is true about the reaction in the conditions of part 3?

A The reaction is at equilibrium B Cannot be determined C The reaction is NON spontaneous D The reaction is spontaneous

Part 3: Which represents the temperature dependance of the spontaneity of this process?

A The reaction is NEVER spontaneous B The reaction is spontaneous at LOW temperatures only C The reaction is spontaneous at HIGH temperatures only D The reaction is ALWAYS spontaneous

Part 4: At what temperature (in K) will the reaction change between being spontaneous and non-spontaneous? If the reaction is NEVER or ALWAYS spontaneous, enter a '0' below.

Ammonia is formed by the Haber process according to the following reaction N2(g) + 3H2(g) 2NH3(g) Use the following data table to answer the questions below Substance: N2(g) H2(g) NH3(g) S° (I/(mol K) 193.7 129.2 194.8 0 0 46.7

Explanation / Answer

Hrxn = sum of H of products- sum of H of reactants

Hrxn = [2 X HNH3]- [HN2 + 3HH2]

Hrxn = [2 X (-46.7]-[0 +0] = -93.4kJ/ mole

(ii) Srxn = sum of S of products- sum of S of reactants

Srxn = [2 X SNH3]- [SN2 + 3SH2]

Srxn = [194.8 X 2]-[193.7 + 3 X 129.2] = -191.7 J / mol K = -0.1917 KJ / mol K

G = H - TS

G = -93.4 - (298 X (-0.1917)) = -93.4 - (-57.127) = -36.27 kJ / mole

Reaction is spontaneous as G < 0

b) The reaction is spontaneous at low temperature as H and S both are negative.

c) G = 0 (at equilibrium) = H - TS = -93.4 - (T X (-0.1917))

93.4 = 0.917 T

T = 487.21 (below this temperature the reaction is spontaneous and above this temperature reaction is non spontaneous.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---