Question 2 Consider the reaction 2 NO(g) + Br2(g) = 2 NOBr2(g) Kc = 1.3 x...
QUESTION 13 [CLO-5] At 1000 K, the equilibrium constant for the reaction 2NO (g) + Br2 (g) ==== 2NOBr (g) is Kp = 0.013. Calculate Kp for the reverse reaction, 2NOBr (g) 2NO (g) + Br2 (9) ==== -0.013 1.1 77 0.013 0 1.6x 10-4
The value of the equilibrium constant for the reaction 2HBr(g)<....>H2(g)+Br2(g) is Kc=1.26*10^-12 at 500k. A. what would be the value of the equilibrium constant K'c for the related reaction written in the following fashion? 1/2 H2(g)+1/2 Br2(g)<....>HBr(g). B. what will be the corresponding value for Kp, the pressure form of the equilibrium constant? (R=0.08206)
1) Consider the following reaction at equilibrium: H2(g) + Br2(g) = 2 HBr(g) Kc = 3.8 x 104 a) Is this reaction reactant-favored or product-favored? (1 point) Answer: b) Based on the given equilibrium reaction, determine the value of the equilibrium constant for the following reaction: Show your work for full credit! (4 points) 2 HBr(g) = H2(g) + Br2(g) Kc = c) Use both the equation and your answer from Part b to answer the problem. In a 1.00...
12. The equilibrium constant for the reaction: SO2(g) + O2(g)so,(g) is Kc # 1.3 x 10, What is the equilibrium constant for the reaction: 2S0,(g)O,(g)+ 2S02(g) 13. For the following reaction, Kc (@ 500 K) 1.0 x 10 2 SO-(g) + 02(g) 2 SO,(g). What is the value of Kp for this reaction at 500 K? (3 Pts.)
Question 17 5 pts Consider the reaction 2 NO(g) + Br2(g) =2 NOBr(g) Kp = 0.0352 (at 298K) What is the Kc for the reaction at the same temperature (298K)? 0.861 10.5 0.0352 1.44 10-3
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2. A) For the reaction 3A(g)+3B(g)⇌C(g) Kc =...
For the equilibrium reaction, 2 S03 - 2 SO2 (g) + O2 (g), Kc is 4.08 x10-3 at 1000 K. Calculate the value for Kp. (R 0.0821 L.atm/ mol.K) A. 9.60 B. 2.99 OC. 4.97 x 10-5 OD.0.335
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2. For the reaction 2A(g)+2B(g)⇌C(g) Kc = 80.2...
1. The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2. Part A For the reaction 3A(g)+2B(g)⇌C(g)...
The equilibrium constant for the reaction: 2NO(g) + Br2(g) <----> 2NOBr(g) is Kc = 1.3x10^-2 at 1,000 Ka.) At this temperature, does the equilibrium favor the product or reactants?b.) Calculate Kc for 2NOBr <----> 2NO + Br2c.) Calculate Kc for NOBr <----> NO + 1/2Br2