A.) Calculate K for a reaction that has ∆H0 = -5.65 KJ/mol and ∆S0 = 43.75 J/mol K at 298.15 K.
B.) Calculate ∆S0 for a reaction that has ∆H0 = 11.36 KJ/mol and K = 0.000376 at 298.15 K. Report your answer in J/mol K
C.)
Consider the reaction below.
2 NH3 ⇋ N2 + 3 H2
When 1.45 M NH3 wis allowed to decompose until equilibrium has been achieved, 0.87 M NH3 remains. If this reaction has ∆S0 = 69.27 J/mol K, calculate ∆H0 for this reaction. Report your answer in J/mol.
A.) Calculate K for a reaction that has ∆H0 = -5.65 KJ/mol and ∆S0 = 43.75...
What is the ΔrG for the following reaction (in kJ mol-1) at 298 K? N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) The conditions for this reaction are: PN2 = 1.47 bar PH2 = 0.35 bar PNH3 = 1.45 bar
Find ΔrG for the following (in kJ mol-1) N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) The conditions for this reaction are: Temp: 298k P - NH3 = 0.95 bar P - H2 = 1.95 bar P - N2 = 1.25 bar NH3(g) ?H ∙(kJ mol-1) = -45.9 ?G ∙(kJ mol-1) = -16.4 S ∙(J K-1 mol-1)192.8 N2(g) ?H ∙(kJ mol-1) = 0 ?G ∙(kJ mol-1) = 0 S ∙(J K-1 mol-1)191.6 H2(g) ?H ∙(kJ mol-1) = 0...
4. The standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-' at 298 K. Given the heat capacity data below and the data in Problem 2, calculate the standard enthalpy of formation at 1200 K Cp.m (H2 (9))/ J mol K-1 = 29.1 - (0.84 x 10- K-)T Cpm (N2 (g))/ J mol K-1 = 26.98 +(5.9 x 10-'K-!)T
Calculate the enthalpy change for the following reaction using bond enthalpies in kJ/mol. *Note: Balance first! N2 + H2 → NH3
Question 4 1 pts Calculate ASºsys for this reaction at 25°C. Input final answer in units of J/K and to 1 decimal place. N2(g) + 3 H2(g) → 2 NH3(g) Species Sº (J/mol K) AH°F (kJ/mol) N2(g) 191.5 o H2(g) 130.6 NH3(g) 192.8 -45.92
QUESTION 12 Given that AG for NH=-16.667999999999999 kJ/mol, calculate the equilibrium constant for the following reaction at 298 K: N2(8)+3 H2()2 NH3(g) Oa5.820000000000003 x 108 b8 349999999999996 x 102 Oc1.01 Od696999 998 x 10s O4.5099999999999998 x <1069
A reaction has and △H°298 = 106 kJ/mol and △S°298 = 326 J /mol K at 298 K. Calculate △G in kJ/mol.
Nitrogen AHºf (kJ/mol) AG°f (kJ/mol) Sº (J/mol K) N2 (g) 0 0 191.6 472.7 455.6 153.3 N(g) NH3 (g) -46.1 -16.5 192.5 NH3 (aq) -80.0 -27.0 111.0 NH4+ (aq) -132.0 -79.0 113.0 NO (g) 90.3 86.6 210.8 NOCI (g) 51.7 66.1 261.8 NO2 (g) 33.2 51.3 240.1 N20 (g) 82.1 104.2 219.9 N204 (8) 9.2 97.9 304.3 Consider the reaction N2(9) + 202(9)—2NO2(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K...
ΔH0 (kJ/mol) C2H2 (g) = 226.7 C6H6 (g) = 82.9 S0(J/mol*K) C2H2 (g) = 200.8 C6H6 (g) = 269.2 Cp C2H2 (g) Cp = 30.7 + 5.28*10-2 T + 1.63*10-5 T2 (J/mol*K) C6H6 (g) Cp = -1.7 + 3.25 *10-2 T - 11.06*10-5 T2 (J/mol*K) Compute ΔH(Temperature Final)reaction and ΔS(Temperature Final)reaction Initial Temperature = 298.15K Final Temperature = 330K
The following reaction has the thermodynamic values at 298 K: AH° =-136.9 kJ/mol and AS" = -120.6 J/mol K. HаС — СHз (9) Нас— CHz (0) + Hа (0) a. Calculate AG° at 298 K for this reaction in kJ/mol (Enter your answer to four significant figures.) kJ/mol b. The reaction is c. The reaction is.