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 = 80.6 at a temperature of 351 ∘C
Calculate the value of Kp.
For the reaction
X(g)+3Y(g)⇌2Z(g)
Kp = 3.00×10−2 at a temperature of 83 ∘C.
Calculate the value of Kc
5. The equilibrium constant, KcKc, is calculated using molar concentrations. For gaseous reactions another form of...
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)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...
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)?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...
, 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...
Part A For the reaction 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 K = K (RT)An where R=0.08206 L-atın/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 the reaction N (g)...
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...
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at 625 oC What is Kc at this temperature? Kp = Kc[RT]Δn R = 0.08206 L-atm/mol K