For H2O(l) ? H2O(g) ?rG298K = 8.596 kJ/mol and ?rH/j mol-1 =
57086.5 -45.327(T/K) + 4.966 x 10-3 (T/K)2 Compute ?rG380K
For H2O(l) ? H2O(g) ?rG298K = 8.596 kJ/mol and ?rH/j mol-1 = 57086.5 -45.327(T/K) + 4.966...
Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
the enthalpy of combustion of CH4(g) to make H2O(l) and CO2(g) is -2340 kJ mol-1. The enthalpy of combustion of CH2(g) to make H2O(l) and CO2(g) is -2760 kJ mol-1. The enthalpy of formation of H2O(l) is -286 kJ mol-1. All the data are for 298 K. The heat capacities for O2(g), CHA(8), CH3(g), H2O(l) and CO2(8) are 29, 61, 71, 75 and 37 JK"mor", respectively. Deduce a) 4U298 for the combustion of C4H8(g). 5) AH for the combustion of...
termine AHo for the combustion of ethanol. Given: AHof (in kJ mol-1): C2H5OH(): -277.69 CO2(g): -393.51 H2O(l): -285.83 • 13. Calculate AHO, 298K for the following reaction: CO(g) + 1/2 02(g) 2 CO2(g) Given: A Hof,298K (in kJ mol-1): CO(g): -110.53 CO2(g): -393.51
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
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...
The following table lists some enthalpy of formation values for selected substances. Substance ΔfH∘ΔfH∘ (kJ mol−1)(kJ mol−1) CO2(g)CO2(g) −393.5−393.5 Ca(OH)2(s)Ca(OH)2(s) −986.1−986.1 H2O(l)H2O(l) −285.8−285.8 CaCO3(s)CaCO3(s) −1207−1207 H2O(g)H2O(g) −241.8−241.8 Part A: Determine the enthalpy for this reaction: Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l) C a ( O H ) 2 ( s ) + C O 2 ( g ) → C a C O 3 ( s ) + H 2 O ( l ) Express your answer in kJ mol−1 k J m o l...
1. Consider a process at 286 K with ΔH = -72.5 kJ/mol and ΔS = -96.8 J/K mol Which of the following is true? ΔG = -44.8 kJ/mol and the process will not be spontaneous ΔG = +39.4 kJ/mol and the process will be spontaneous ΔG = -44.8 kJ/mol and the process will be spontaneous ΔG = -19.1 kJ/mol and the process will not be spontaneous ΔG = +39.4 kJ/mol and the process will not be spontaneous 2. Consider a...
2 H2S(g) + 3 O2(g) → 2 H2O(l) + 2 SO2(g) Compound ΔHof (kJ/mol) Sof (J/mol.K) H2S(g) -20.6 205.8 O2(g) 0 205.2 H2O(l) -285.8 70.0 SO2(g) -296.8 248.2 1.Calculate ΔHrxn. 2.Calculate ΔSsurroundings. 3Calculate ΔSuniverse.
Given the following data, 2 O3(g) ----> 3 O2(g) ∆rH = -377.00 kJ mol-1 O2(g) ----> 2 O(g) ∆rH = 545.00 kJ mol-1 NO(g) + O3(g) ----> NO2(g) + O2(g) ∆rH = -124.00 kJ mol-1 Determine ∆rH for the reaction, NO(g) + O(g) ----> NO2(g) ∆rH = ? kJ mol-1
1.For the reaction
CH4(g) +
H2O(g)3H2(g)
+ CO(g)
H° =
206 kJ and S° =
215 J/K
G° for this
reaction would be negative at temperatures (above,
below) K.
For the reaction
2.2Fe(s) +
3Cl2(g)2FeCl3(s)
H° =
-799 kJ and S° =
-440 J/K
G° would be
negative at temperatures (above, below) K.
4.Consider the reaction:
NH4NO3(aq)N2O(g)
+ 2H2O(l)
Using standard absolute entropies at 298K, calculate the entropy
change for the system when 2.27
moles of NH4NO3(aq) react at
standard conditions.
S°system...