problem #9 Calculate the equilibrium constant for the following reaction at 25°C. 2H2(g) + O2(g) 2H20cv)...
Calculate the standard entropy change for the reaction 2H2(g)+O2(g)?2H2O(l) using the data from the following table: Substance ?H?f (kJ/mol) ?G?f (kJ/mol) S? [J/(K?mol)] H2(g) 0.00 0.00 130.6 O2(g) 0.00 0.00 205.0 H2O(l) -285.8 -237.2 69.90 Express your answer to four significant figures and include the appropriate units. Please show me the steps on how to solve this!!! Thank you!!!
6. Lightning bolts have been found to enable the following atmospheric reaction: N26) + O2(g) → 2 NO(g) 4,Gº = 175.2 kJ mol-1 at 298 K a) (2 marks) The AH of this reaction is 182.6 kJ mol- and the A.Sº is 24.8J molK1. Calculate the temperature at which this reaction changes between being spontaneous to non-spontaneous. Report your answer in K. Show your work b) (3 marks) Using the reaction above, along with the knowledge that A,Gº (NO2 )...
Given the following information at 25°C, Calculate AS° for the h drogenas ethylene: C2H2(g) + 2H2(g) → C2H6(g) Sº Species J/(mol.K) H2(g) C2H2(g) C2H6(g) 130.68 200.94 229.60 a) 561.22 J/K Ob) 232.70 J/K Oc) -232.70 J/K d) -561.22 J/K
The equilibrium constant Kp for the reaction 2H2O(g)--> 2H2(g)+O2(g) is 2x10^-42 at 25 degrees C. (a) what is Kc for the reaction at the same temperature? (b) the very small value of Kp (and Kc) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.
A. Calculate A Gº and Kp for the following equilibrium reaction at 25°C. The AG, values are 0 for Cl2 (9), -286 kJ/mol for PC13 (9), and -325 kJ/mol for PC15 (9). B. Now calculate AG for the reaction if the partial pressures of the initial mixture are PPC15=0.0029 atm, PPC13=0.27 atm, and PC12=0.40 atm Attach File
2. Calculate K for the reaction: IC (g) at 298 K CH,(g) + 3Cl2(g) = CHCI;(8) + 3HCI(g) at 2 A/Gº(CH.(g)) = -50.72 kJ/mol A,Gº(CHCI (g)) = -71.1 kJ/mol AG°(HCl(g)) --95.30 kJ/mol 10-14 at 25 °C. Find the value of K, 3. For the following chemical equilibrium, Kp = 4.6 x 10-14 at 25 for this reaction at 25 °C. 2012(g) + 2 H2O(g) = 4 HCI(g) + O2(g)
1. Calculate A Gº for the reaction at 25°C 2 SO2(g) + O2(g) → 2 SO3 (g), given that AHº is -197.7 kJ mol- and A.Sº is -188 JK' mol!
The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)⇌CH3OH(g)
Part A Calculate Kp for the reaction below. CH3OH(g) CO(g)+2H2(g) Submit My Answers Give Up Part B Reactants will be favored at equilibrium. O Products will be favored at equilibrium. Submit My Answers Give Up Part C Calculate Kp for the reaction below. 를 CO (g) + H2 (g)- CH, OH (g) K=
Calculate the equilibrium constant for the following reaction at 25 °C and 150°C. 11. (2) NH CI() NH()HC1(g) NH,C1 (s) -202.87 NH( -16.45 HC1(8) -95.30 AG (kJ-mo1-1) AG [-16.45(-95.0 -202.87] 91.12 kJ At 298 K: K#e*(- Gran/RT) e*(91120/(8.314 x 298) 1.1 At 423 K: K= e"(-aGrin/RT)-e*(91 120/(8.314 x 423) = 5.6 × 10-16 10-12
Calculate the equilibrium constant for the following reaction at 25 °C and 150°C. 11. (2) NH CI() NH()HC1(g) NH,C1 (s) -202.87 NH( -16.45 HC1(8) -95.30 AG...
1 What is the equilibrium constant for a reaction at temperature 89.1 °C if the equilibrium constant at 22.6 °C is 49.93? For this reaction, ΔrH = -21.1 kJ mol-1 . 2 What is the ΔrG° for the following reaction (in kJ mol-1)? C6H12O6(s, glucose) + 6 O2 (g) ⇌6 CO2 (g)+ 6 H2O (l) 3 What is the ΔrG° for the following reaction (in kJ mol-1)? 2 NO2 (g) ⇌N2O4 (g) 4 What is the ΔrG for the following...