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Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 839...
Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 641...
Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 641 K for the following reaction. NH4Cl(s) NH3(g) + HCl(g) ΔΗ /kJ mol-1 -314.4 -45.9 -92.3 Smº /JK-mol-1 94.6 192.8 186.9 Cp.m /JK-mol-1 84.1 35.1 29.1 Do not use the Van't Hoff equation, In(K/K) = -(AHR/R) (T2-1-T1-1) The value of the thermodynamic equilibrium constant is Kp = Number
Using standard thermodynamic data (linked), calculate the equilibrium constant at 298.15 K for the following reaction....
Using standard thermodynamic data (linked), calculate the equilibrium constant at 298.15 K for the following reaction. HCl(g) + NH3(g)--->NH4Cl(s) K =
Use standard thermodynamic data (in the Chemistry References) to calculate AG at 298.15 K for the...
Use standard thermodynamic data (in the Chemistry References) to calculate AG at 298.15 K for the following reaction, assuming that all gases have a pressure of 10.88 mm Hg. 2NO(g) + O2(g)—>2NO2(g) AG= kJ/mol Nitrogen AHºf (kJ/mol) AG°f (kJ/mol) sº (J/mol K) N2(g) 191.6 N(9) 472.7 455.6 153.3 NH3(9) -46.1 -16.5 192.5 NH3(aq) -80.0 -27.0 111.0 NH4+ (aq) -132.0 -79.0 113.0 NO(9) 90.3 86.6 210.8 NOCI(g) 51.7 66.1 261.8 NO2(9) 33.2 240.1 51.3 104.2 N2O(9) 82.1 219.9 N204(9) 9.2 97.9...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.18...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
Calculate the value of Kp for the reaction $$2N2(g)+O2(g) 2N2O(g) at 298.15 K and 1273 K....
Calculate the value of Kp for the reaction $$2N2(g)+O2(g) 2N2O(g) at 298.15 K and 1273 K. Thermodynamic data for N2O(g) are: ΔH°f = 82.05 kJ/mol; S° = 219.9 J/mol ·K; ΔG°f = 104.2 kJ/mol. Pt 1: 298.1K Pt 2: 1273 K
Given the information below, calculate ∆?∘?,298.15? and ∆?∘?,400? for the following reaction: ??(?) + 2???(?) →...
Given the information below,
calculate ∆?∘?,298.15? and ∆?∘?,400? for the following
reaction:
??(?) + 2???(?) → ?2(?) + ????2(?)
Assume that heat capacities are constant over the desired
temperature range. All molar enthalpies of formation are at
298.15K.
∆?∘ Mg(g) ? = 147.1 kJ/mol
∆?∘ HCl(g) ? = -92.3 kJ/mol
∆?∘ H2(g) ? = 0 kJ/mol
∆?∘ MgCl2(s) = -641.3 kJ/mol
Cp,m Mg(g) = 20.8 J mol-1 K-1
Cp,m HCl(g) = 29.1 J mol-1 K-1
Cp,m H2(g) = 28.8 J...
Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the...
Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 14.96 mm Hg. 2N2(g) + O2(g)2N2O(g) G =_____ kJ/mol
Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the...
Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 18.65 mm Hg. 2NO(g) + O2(g)2NO2(g) G = ____kJ/mol
a.) Calculate the equilibrium constant for the following reaction at 298.15 K from cell potential data....
a.) Calculate the equilibrium constant for the following reaction at 298.15 K from cell potential data. Express the answer as lnK. Sn4+ + 2Fe2+ ----> Sn2+ + 2Fe3+ b.) Calculate the standard Gibbs free energy change in kJ/mol at 298.15 K for the following reaction from cell potential data: 3Sn4+ + 2Cr ----> 3Sn2+ + 2Cr3+
Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 641 K for the following reaction. NH4Cl(s) NH3(g) + HCl(g) ΔΗ /kJ mol-1 -314.4 -45.9 -92.3 Smº /JK-mol-1 94.6 192.8 186.9 Cp.m /JK-mol-1 84.1 35.1 29.1 Do not use the Van't Hoff equation, In(K/K) = -(AHR/R) (T2-1-T1-1) The value of the thermodynamic equilibrium constant is Kp = Number
Use standard thermodynamic data (in the Chemistry References) to calculate AG at 298.15 K for the following reaction, assuming that all gases have a pressure of 10.88 mm Hg. 2NO(g) + O2(g)—>2NO2(g) AG= kJ/mol Nitrogen AHºf (kJ/mol) AG°f (kJ/mol) sº (J/mol K) N2(g) 191.6 N(9) 472.7 455.6 153.3 NH3(9) -46.1 -16.5 192.5 NH3(aq) -80.0 -27.0 111.0 NH4+ (aq) -132.0 -79.0 113.0 NO(9) 90.3 86.6 210.8 NOCI(g) 51.7 66.1 261.8 NO2(9) 33.2 240.1 51.3 104.2 N2O(9) 82.1 219.9 N204(9) 9.2 97.9...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
Given the information below,
calculate ∆?∘?,298.15? and ∆?∘?,400? for the following
reaction:
??(?) + 2???(?) → ?2(?) + ????2(?)
Assume that heat capacities are constant over the desired
temperature range. All molar enthalpies of formation are at
298.15K.
∆?∘ Mg(g) ? = 147.1 kJ/mol
∆?∘ HCl(g) ? = -92.3 kJ/mol
∆?∘ H2(g) ? = 0 kJ/mol
∆?∘ MgCl2(s) = -641.3 kJ/mol
Cp,m Mg(g) = 20.8 J mol-1 K-1
Cp,m HCl(g) = 29.1 J mol-1 K-1
Cp,m H2(g) = 28.8 J...
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