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F6 Chem Eqm. Constant Kc (2)
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最佳解答:
N2O4(g)D2NO2(g) Initial moles0.0240 Equilibrium moles0.024-α2α Concentrations in equilibrium: [N2O4] = (0.024-α) / 0.372 mole/dm3 [NO2] = (2α) / 0.372 mole/dm3 Kc = 4.61 × 10^-5 = [NO2]^2/[N2O4] = (2α / 0.372)^2 / [(0.024-α) / 0.372] ==> 4α^2 / [ 0.372×(0.024-α) ] = 4.61 × 10^-5 4α^2 - 4.61 × 10^-5 × 0.372 × (0.024-α) = 0 α = 3.186 × 10^-4 (using formula to solve the 2nd-order equation) Check: [N2O4] = (0.024 - 3.186×10^-4) / 0.372 = 6.366 × 10^-2 mole/dm3 [NO2] = 2 × 3.186×10^-4 / 0.372 = 1.713 × 10^-3 mole/dm3 Kc = [NO2]^2 / [N2O4] = (1.713×10^-3)^2 / (6.366×10^-2) = 4.61 × 10^-5 OK! Dissociation rate = 3.186 × 10^-4 / 0.024 × 100% = 1.33%
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F6 Chem Eqm. Constant Kc (2)
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An 0.024 mole sample of N2O4(g) is allowed to dissociate and come to equilibrium with NO2(g) in an 0.372 dm3 vessel at 298K. What is the percentage dissociation of the N2O4(g) if Kc for N2O4(g)D2NO2(g) is 4.61*10^-5 mol dm-3 Ans: 0.0133%最佳解答:
N2O4(g)D2NO2(g) Initial moles0.0240 Equilibrium moles0.024-α2α Concentrations in equilibrium: [N2O4] = (0.024-α) / 0.372 mole/dm3 [NO2] = (2α) / 0.372 mole/dm3 Kc = 4.61 × 10^-5 = [NO2]^2/[N2O4] = (2α / 0.372)^2 / [(0.024-α) / 0.372] ==> 4α^2 / [ 0.372×(0.024-α) ] = 4.61 × 10^-5 4α^2 - 4.61 × 10^-5 × 0.372 × (0.024-α) = 0 α = 3.186 × 10^-4 (using formula to solve the 2nd-order equation) Check: [N2O4] = (0.024 - 3.186×10^-4) / 0.372 = 6.366 × 10^-2 mole/dm3 [NO2] = 2 × 3.186×10^-4 / 0.372 = 1.713 × 10^-3 mole/dm3 Kc = [NO2]^2 / [N2O4] = (1.713×10^-3)^2 / (6.366×10^-2) = 4.61 × 10^-5 OK! Dissociation rate = 3.186 × 10^-4 / 0.024 × 100% = 1.33%
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