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QUESTION 7 For the reaction 2NO2(g)-N204(g), Kat 25°C is 7.3, when all partial pressures are expressed...
2. For the isomerization reaction, N204(g) + 2NO2(g), the degree of dissociation (a) is 0.201 at equilibrium (25°C and 1 bar). (NOTE: DO NOT simply look up the thermodynamic table for this question. You MUST compute all the quantities using known formula.) (a) Calculate A,G,A,Gº and K at equilibrium. (b) If the equilibrium constant at 100°C is 17.46, is this reaction exothermic or endothermic? (c) Calculate the standard reaction entropy, A. Sº, of this reaction.
10. The equilibrium constant for the reaction N204(8) = 2NO2(g) at 25 °C is 5.88 x 10-3. Suppose 15.6 g of N204(g) is placed in a 5.000-L flask at 25 °C. Calculate the following: (a) The moles of NO2(g) present at equilibrium (b) The percent dissociation of the original N,O4(g).
Consider the following reaction: 2NO2(g) ⟶ N2O4(g) Part B Calculate AG at 298 K if the partial pressures of NO2 and N204 are 0.39 atm and 1.62 atm , respectively. Express the free energy in kilojoules to two decimal places. PO AQ R o aj ? AG = Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining
At 25°C, the equilibrium partial pressures for the reaction 3 A(g) + 4B(g) = 2(g) + 3D(8) were found to be PA = 4.26 atm, PB = 5.44 atm, Pc = 4.00 atm, and Pb = 5.30 atm. What is the standard change in Gibbs free energy of this reaction at 25°C? AGеxn = AGran = 0.44 0.44 kali
At 25 ∘C , the equilibrium partial pressures for the reaction 3A(g)+3B(g)↽−−⇀C(g)+3D(g) were found to be ?A=4.55 atm, ?B=4.49 atm, ?C=5.77 atm, and ?D=5.62 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C ? (Answer in kj/mol)
At 25 ∘C , the equilibrium partial pressures for the reaction 3A(g)+4B(g)↽−−⇀2C(g)+2D(g) were found to be ?A=5.80 atm, ?B=5.34 atm, ?C=5.59 atm, and ?D=4.57 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘C ? Δ?∘rxn= ______ kJ/mol
Consider the reaction A(g) + 2B(g) ⇌ C(g) at 25 °C. The initial pressures of all the gases are 0.100 atm. Once equilibrium has been established, it is found that Pc = 0.140 atm. What is Kp for this reaction at 25 °C. Please explain and/or show the steps.
For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3. Calculate [NO2] at equilibrium when [N2O4]=6.90×10−2mol/L.
The generic reaction below has Kp = 25 when the temperature is 225°C: C(g) ⇌ D(g) If the initial partial pressure of C (PC) is 5.0 atm and the initial partial pressure of D (PD) is 0.0 atm, what are the partial pressures of the two gases at equilibrium? Question 7 options: PC = 0.6 atm and PD = 4.4 atm PC = 4.4 atm and PD = 0.60 atm PC = 1 atm and PD = 25 atm PC =...
please Help its due in 1 hour 3. Consider the following reaction at 25°C N204(g) = 2NO2(g) AG° = 5.40 kJ ontains initial concentrations of 0.453 M N2O4 and 0.150 M NO2. What A reaction vessel contains initial concentrang is the AG for the reaction at this temperature? a. -7437 kJ b. -68.00 kJ c. 9.08 kJ d. -2.04 kJ e. 5.32 kJ A mixture of 0.500 mol H2 and 0.500 mol l was placed in a 1.00 L vessel...