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calculate k at 298 k for the following reaction: 2 no(g) cl2(g) ⇌ 2 nocl(g) × 10 (enter your answer in scientific notation.) standard thermodynamic values at 298 k substance δg o f (kj/mol) no(g) 86.60 nocl(g) 66.07 no2(g) 51 n(g) 456 cl(g) 105.0
calculate k at 298 k for the following reaction: 2 no(g) cl2(g) ⇌ 2 nocl(g) × 10 (enter your answer in scientific notation.) standard thermodynamic values at 298 k substance δg o f (kj/mol) no(g) 86.60 nocl(g) 66.07 no2(g) 51 n(g) 456 cl(g) 105.0
At 298 K, theequilibriumconstant (K) for thereaction:2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) is approximately 278.192To calculate the equilibrium constant (K) at 298 K for thereaction2 NO(g) + Cl2(g) ⇌ 2 NOCl(g), we need to use the standard Gibbs free energy of formation (ΔG°f) values for the substances involved.The equation for calculating K is as follows:K = exp(-(ΔG°) / (RT))Where:ΔG° = Σ(nΔG°f products) - Σ(nΔG°f reactants)R = Gasconstant(8.314 J/(mol·K))T = Temperature in Kelvin (298 K)Let's calculate K using the provided ΔG°f values:ΔG° = [2(ΔG°f NOCl) - (ΔG°f NO) - (ΔG°f Cl2)]= [2(66.07) - 86.60 - 0] = -35.06 kJ/molNow we can substitute the values into theequation:K = exp(-(-35.06 × 10^3) / (8.314 × 298))Calculating the exponential term:K ≈ exp(13920.68 / 2470.472)K ≈ exp(5.633)Finally, evaluating the exponential function:K ≈ 278.192 (approximately)Therefore, at 298 K, the equilibrium constant (K) for the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) is approximately 278.192 (in scientific notation, 2.78192 × 10^2).Learn more aboutreaction:brainly.com/question/16737295#SPJ11...