The equilibrium constant for the reaction SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) is given by the expression Kc = [SO3][NO]/[SO2][NO2], where square brackets denote molar concentrations. The numerical value of this equilibrium constant would depend on the specific conditions of the reaction.
The amount of NO2 and SO2 eould
The equilibrium constant expression for the reaction is Kc = [H2O]^2/[SO2][O2]. Given the concentrations at equilibrium, we can solve for [H2O]. Plugging in the values, we get 31.25 = [H2O]^2 / (0.03)(0.05). Solving for [H2O] gives us [H2O] = sqrt(31.25 * 0.03 * 0.05), which is approximately 0.275M.
SO2 stands for sulfur dioxide and NO2 stands for nitrogen dioxide.
Reduction reaction: 2NO2 + 2H+ + 2e- -> N2O4 + H2O Oxidation reaction: 2NO2 -> N2O5 + e-
The equilibrium constant for the reaction SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) is given by the expression Kc = [SO3][NO]/[SO2][NO2], where square brackets denote molar concentrations. The numerical value of this equilibrium constant would depend on the specific conditions of the reaction.
SO2(g) + NO2(g) ==> SO3(g) + NO(g)Keq = [SO3][NO]/[SO2][NO2] Without knowing concentrations, one cannot calculate the actual value of Keq.
SO2(g) + NO2(g) ==> SO3(g) + NO(g)Keq = [SO3][NO]/[SO2][NO2] Without knowing concentrations, one cannot calculate the actual value of Keq.
The amount of NO2 and SO2 eould
The equilibrium constant expression for the reaction is Kc = [H2O]^2/[SO2][O2]. Given the concentrations at equilibrium, we can solve for [H2O]. Plugging in the values, we get 31.25 = [H2O]^2 / (0.03)(0.05). Solving for [H2O] gives us [H2O] = sqrt(31.25 * 0.03 * 0.05), which is approximately 0.275M.
The amount of NO and SO3 would increased.
SO2 stands for sulfur dioxide and NO2 stands for nitrogen dioxide.
keq= [SO2]2[O2]/[SO3]2
Reduction reaction: 2NO2 + 2H+ + 2e- -> N2O4 + H2O Oxidation reaction: 2NO2 -> N2O5 + e-
SO3 + NO SO2 + NO21.7M + xM |=| 0.070M + 1.3MKeq = 10.8 = [SO2]*[NO2] / [SO3]*[NO] = 0.070*1.3 / 1.7*xSolving x (concentration NO at eq.) 10.8 * (1.7*x) = 0.070*1.3x = 0.0091 / 18.36 = 0.000496 = 0.00050 M = 0.50 mM
NO2 and SO2
Nitrogen and Sulphur gases such as NO2 and SO2 are affecting Taj Mahal in getting yellowish.