The complete calculation of the temperature is given as below:-
4.57. For the equation 2H20 (g) 2H2 (g)+ 02 (g) -241.8 kJ and AG(25°C) = -228.61 kJ. Use the AH(25°C) Gibbs-Helm...
Part C Given the following reactions 2H2 (9) + O2(g) → 2H20 (9) AH = - 483.64 kJ 2H20 (1) ► 2H2(g) + O2 (9) AH = 571.66 kJ Calculate the enthalpy for the condensation of water vapor to liquid, given by this reaction: H2O(g) + H20 (1) +44.01 kJ +88.02 kJ 0 -44.01 kJ -88.02 kJ Submit Request Answer
6) Given the following reactions H20 (1) H20 (8) AH = 44.01 kJ AH = -483.64 kJ 2H2 (8) + O2(g) +2H20 (g) the enthalpy for the decomposition of liquid water into gaseous hydrogen and oxygen 2H20 (1) +2H2 (g) + O2(g) _kJ.
HQ11.43 Homework Answered For the reaction of H2(g)+1/2 02(g) H20(g), AH° is -285.82 kJ. What is AHo for the reaction 2H20(g) -- > 2H2(g) + 02(g)?| HQ11.45 Homework Unanswered Using the first equation in HQ11.43, how grams of 02(g) will be needed to produce 55.0 kJ of heat if reacted in excess H2(g)? Report answer in grams without units Numeric Answer:
QUESTION 7 AH® - -560 kJ Look at this reaction: 2H2(g) + O2(9) --> 2H20(1 Which one of the following statements is true? Heat is absorbed during the reaction The reaction is endothermic The enthalpy change is positive This reaction is exothermic
1. Given AH (CO2(g)) - -393.5 kJ/mole, AH (CH2(g)) = -146.4 kJ/mole HO )) - -241.8/mote and AH (a) (10 points) How much heat is evolved if 36.0 g CsHis undergoes combustion at constant pressure? (b) (10 points) What volume of O2(g) is required in the combustion at 25°C and 1.00 atm if 9,818 kJ are evolved?
298.15K AH HO (1) -285.8 H2O(g) -241.8 4,6° Ch -237.1 -228.6 Sº Go Go 70.0 75.3 188.8 33.6 a) (1 pt) Calculate Acondensation Gº in J) for the condensation of water vapor at 25°C and 1 bar pressure. b) (1 pt) Is this process spontaneous under these conditions? c) (2 pts) Calculate Kp at 25°C. d) (15 pts) Use the Gibbs-Helmholtz equation AG d ) = -AH°(T) T 12 to calculate Acondensation Gº (in J) for the condensation of water...
Gibbs free energy change AG is defined as AG= AH-TAS. For a spontaneous process, AG<0. For a. reversible reaction at equilibrium, AG = 0. The equilibrium constant K of the reversible reaction is relate- AG"=-RTIn(K). Symbol 40" refers to thermodynamics standard condition: 298 K and 1 atm. Exercise:consider the following reaction: bo:pfoiex 2 HNO:(aq)+NO(g)- 3 NO2(g) + H2O(I) AH=+136.5 kJ; AS = +287.5 J/K a. Below what temperature does the following reaction becomes nonspontaneous? niliod sis anotsole s ob io...
NEED ANSWERS ASAP PLZ QUESTION 4 AH° = -560 kJ Look at this reaction: 2H2(g) + O2(g) --> 2H20(1) Which one of the following statements is true? O Heat is absorbed during the reaction O The reaction is endothermic O The enthalpy change is positive O This reaction is exothermic QUESTION 3 If AH = 25 J for a certain process, that process O releases heat O is exothermic. O is endothermic. O can't tell
Question 5 Correct Use AG° =AH° -TAS° to calculate AG (in kJ) at 298 K for : Mark 1.00 out of 1.00 2002(g) +4H2O(1) ► 2CH3OH(1) + 302(g) P Flag question Answer: 1404.84 Question 6 Not answered Marked out of 1.00 If the above reaction could be done at 3399 K, what would be your estimate for AGº (in kJ) at this elevated temperature? Use AGⓇ =AH°-TAS and assume AH° and AS are independent of temperature. P Flag question (Theis...
AG°= AH-TAS AG=AGº+RTinQ where R=8.314 J/mol K 1. Calculate AGº for the following reaction at 25 °C if AH°= -1854 kJ/mol; AS°= -236 J/mol K CH-COCH3(g) + 402(g) → 3C02(g) + 3H2O(1) 2. NH.NO, dissolving in water is a spontaneous process. As it dissolves, the temperature of the solution decreases. Based on this, what must the signs (positive or negative) of AG, AH, and AS be?