I've been struggling with these homework questions. Any help would be greatly appreciated
1. For the gas phase reaction reaction:
2 NOCl (g) ⇌ 2 NO (g) + Cl2 (g)
[NOCl]eq = 1.98 M
[NO]eq = 0.044 M
[Cl2]eq = 0.032 M
at 35oC.
What is the value of KC at this
temperature?
KC =
2. It is often necessary to use the
thermodynamic equilibrium constant,
K.
K is related to KC by the
equation:
K = KC (RT)Δn (also seen as Kp = KC (RT)Δn in your text)
where Δn = moles of gas in the products - moles of gas in the
reactants
R = the gas constant = 0.08206
T = temperature (K)
Calculate K for this reaction at this temperature.
K =
I've been struggling with these homework questions. Any help would be greatly appreciated 1. For the...
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, 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 =...
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...
5. The equilibrium constant, KcKc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, KpKp, 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)R=0.08206 L⋅atm/(K⋅mol), TT is the absolute temperature, and ΔnΔn 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 3A(g)+2B(g)⇌C(g) KcKc...
The equilibrium constant, KcKc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, KpKp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)ΔnKp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol)R=0.08206 L⋅atm/(K⋅mol), TT is the absolute temperature, and ΔnΔn 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)N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2Δn=2−(1+3)=−2. A For the reaction 3A(g)+3B(g)⇌C(g)3A(g)+3B(g)⇌C(g) KcKc...
The equilibrium constant, KcKcK_c, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, KpKpK_p, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)ΔnKp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol)R=0.08206 L⋅atm/(K⋅mol), TTT is the absolute temperature, and ΔnΔnDelta n 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)N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2Δn=2−(1+3)=−2. A.) For the reaction 3A(g)+3B(g)⇌C(g)3A(g)+3B(g)⇌C(g)...
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)?n where R=0.08206 L?atm/(K?mol), T is the absolute temperature, and ?n 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 ?n=2?(1+3)=?2. Part A For the reaction 3A(g)+3B(g)?C(g) Kc...
Chapter 15 Homework Pressure-Based versus Concentration-Based Equilibrium Constants 11 of 41 Review I Constants I Periodic Table The equilibrium constant, K 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 Part A For the reaction 3A(g)3B(g)C(g) Kp = Kc(RT)^n Ke 68.8 ta temperature of 273 C where R 0.08206 L atm/(K.mol), T is the absolute temperature, and...
, 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 Part A Kp = K.(RT)An For the reaction 3A(g) + 2B(g) = C(g) where R=0.08206 L.atm/(K·mol), T is the absolute temperature, and An is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider...
In a reaction vessel @ 310 K, an equilibrium mixture is found to contain: 0.0370 M PCl3, 0.0174 M Cl2 and 1.316 M PCl5. PCl3 (g) + Cl2 (g) ⇋ PCl5 (g) Calculate Kp for the reaction. I did this, and used the equation (1.316 M PCl5) / (.0370 M Cl3 * .0174 M Cl2) = 2044 = 2040 with sig figs. Then because the delta n is -1 because of the coefficients, I used the equation (2040) / (.08206 *...