Kc=5.85x10^-3 at 25 degrees C for the reaction
fifteen(15.0) grams of N2O4 is confined in a 5.00-L flask at 25 degrees C. Calculate(a) the number of moles of NO2 present at equilibrium and(b) the percentage of the original N2O4 that is dissociated.
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Can you please check if I came up with the right answer for this please. A 4.30L container had 55.10 moles of N2O4 added to it. When equilibrium was established the concentration of N2O4 was found to be 1.69M.What is the Kc for
92.01 grams of N2O4 (g) is placed in a container and allowed to dissociate. N2O4 (g) --> 2NO2 (g) The mixture of N2O4 and NO2 resulting from the reaction occupies 36.0 liters at a total pressure of 773 mmHg and 45 °C. A. Let x
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4 (g) ---> 2NO2 (g) Delta H rxn: 55.3 kJ At 298 K a reaction vessel initially containing .100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2.
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g)ΔHorxn=55.3kJ At 298 K, a reaction vessel initially contains 0.100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What
Calculate the rate at which N2O4 is formed in the following reaction at the moment in time when NO2 is being consumed at a rate of 0.0521 M/s. 2NO2(g) N2O4 (g) I am not really sure how to go about this problem. rate = k (N2O4)
N2O4 --->2NO2 a 1 liter flask is charged with .4 mols of N2O4. at equilibrium at 373 k, .0055 mols of N2O4 remain. what is the Kc for this reaction? please explain to me how to get the rate law for this and solving
Given: N2O4 (g) « 2NO2 (g) @ 25 degrees celcius, Kc is 5.84 x 10^-3. (A) Calculate the equilibrium concentrations of both gases when 4.00 grams of N2O4 is placed in a 2.00 L flask at 25 degrees celcius. (B) What will be the new
What is the process to calculate ΔH° for the reaction? O2(g) + 2NO(g) --> N2O4(g) O2(g) + 2NO(g) --> 2NO2(g) ΔH°1 N2O4(g) --> 2NO2(g) ΔH°2 a) ΔH°1 + ΔH°2 b) ΔH°1 - ΔH°2 c) ΔH°2 - ΔH°1 d) ΔH°1 - 2ΔH°2 I know