解析 解 电极反应 Cu(acu )+e=Cu(s) \varphi_(Cu)⋅ρ_(Cu)=0.521V Cu2+(acu2)+2e-=Cu(s) e \varphi_(Cu)^2⋅x_(i1)=0.337V E =2-2=-2/=0.521-0. 337 =0.184(V) △G=-nFE^-=-2.303RTlgK^0 lgK^θ=(2*9648*0.184)/(2.303*8.314*298)=6.22 K^θ=1.66*10^6 ...
To find the equilibrium constant Kc for the reaction: K2CO3(aq)+BaSO4(s)⇌BaCO3(s)+K2SO4(aq) we will follow these steps: Step 1: Write the expression for Kc The equilibrium constant Kc is defined as the ratio of the concentrations of the products to the concentrations of the reactan...
Write the equilibrium constant (Kc) expression for the following reaction : (i)Cu2+(aq)+2A(s)⇔Cu(s)+2A+(aq) (ii)4HCI(g)+O2(g)⇔2CI2(g)+2H2O(g) View Solution Write the equilibrium constant expressions for the following reactions : (i)12N2(g)+32H2(g)⇔NH3(g) (...
最好是不要直接给答案,麻烦写下,Q10 The equilibrium constant K p for the dissociation reaction of the
A. reactants B. products C. equilibrium state D. reverse reaction 相关知识点: 试题来源: 解析 B。平衡常数大说明反应更倾向于生成产物。选项 A“reactants”表示反应物;选项 B“products”表示产物;选项 C“equilibrium state”表示平衡状态;选项 D“reverse reaction”表示逆反应。所以答案是 B。反馈...
+/c=0.521V,2c=0.337V 相关知识点: 试题来源: 解析 解电极反应Cu+(ac+)+e-Cu(s)=0.521VCu2+(ac2+)+2e-==Cu(s)2+cu=0.337VE=-=-2+=0.521-0.337=0.184(V)AG = -nFE= -2. 2×96485×0.1846.222.303×8.314×298K=1.66×10 反馈 收藏 ...
The equilibrium constantKcfor the reaction H2(g)+CO2(g)⇌H2O(g)+CO(g) is4.20at1650.00°C.Initially0.750molH2and0.750molCO2are injected into a2.10Lflask. Calculate the concentration of each species at equilibrium. Be sure...
Tags Constant Equilibrium Equilibrium constant In summary: CO2] = 0.0685 M, [H2O] = 0.210 M, [H2] = 0.0685 MSo in summary, to find the equilibrium constant (Kc) for the reaction CO2 + H2 <-> CO + H2O, with initial concentrations of 0.229 mol CO2, 0.229 mol H2, and 0.328 mol H2O...
36 at 325 K and 100 kPa.What is the percent conversion of N2 O4 ?N_2O_4(g)⇌2NO_2(g))(Colorless)(Brown) 相关知识点: 试题来源: 解析 2.10 K^θ=2.25 ,HI的转化率=65.4% 2.11 Q =1.16,反应逆向进行 2.12 K^θ=0.54,△_rG_m=11.9kJ⋅mol^(-1) ,反应逆向进行 反馈 收藏 ...
The equilibrium constant,K,80×10-2at6 2HI(g)⇌H2(g)+I2(g) An equilibrium mixture of the three gases in a1.00Lflask at698Kcontains0.301MHI,4.04×10-2MH2and4.04×10-2MI.What will be the concentrations of the three ...