divide by stoichiometry(1/2) dN2o5/dt =1/4 dNO2/dt=dO2/dtso dNo2/dt = 4/2*1.60 = 3.2
the rate of formation of NO2 9.12 (mol NO2)?L–1s-1
You need to look at the stoichiometric ratios.Compare the coefficient of N2O5 to NO2. It is 2:4. Or 1:2 if you divide by 2. So that means for every one mole of N2O5 used 2 moles of NO2 will form.So you just need to double the rate given. 1.6 mol/min * 2 = 3.20 mol/min for NO2
Well, one mole of O2 is produced for every 2 moles of N2O5Therefore it must be produced at twice of the rateSo the rate of production of O2 is 2x1.60 mol/min= 3.2 mol/min
Looking at N205, it has 2 mols of N in it so 3.2 mols of N disappears and since we know the amount has to equal on both sides NO2 has to appear at the same rate to remain in equilibrium
rate of appearance of NO2 = twice rate of disappearance of N2O5= 2 * 1.6 mol/min = 3.2 mol/min
Happy To Help You :)
rate of appearence of NO2 = 2*N2O5 = 2*1.6 = 3.2 moles/min
what is the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.11 atm to rise to 0.155 atm?
The decomposition of dinitrogen pentoxide is described by the chemical equation
2 N2O5(g) ? 4 NO2(g) + O2(g)
If the rate of disappearance of N2O5 is equal to 1.60 mol/min at a particular
The following mechanism has been proposed for the decomposition of dinitrogen pentoxide?
It has the experimental rate law, rate = -Delta[N2O5] / Delta t = k[N2O5]: N2O5(g) k2 k1 NO2(g
The decomposition reaction: N2O5 (gas) => N2O4 (gas) + 1/2 O2 (gas) obeys the first order integrated rate law: Rate = k [N2O5] in which k = 1.68 x 10^-2 mol/L-sec at the temperature
The reaction 2N2O5 ? 2N2O4 + O2 obeys the rate law:
rate = k[N2O5], in which the rate constant is 0.00757 s-1, at a certain temperature. If 3.09 moles of N2O5 was placed in a 5.00-liter
[Rate Law Problems]The decomposition of N2O5 in carbon tetrachloride proceeds as follows.
2 N2O5 4 NO2 O2
The rate law is first order in N2O5. At 64°C the rate constant is 4.82 10-3 s-1.
Consider the decomposition of N2O5 in CCl4 at 45 C 2 N2O5 (g)----->2 N2O4 (g) + O2 (g) The rate constant for this order reaction=0.037/min. A sample weighing 68.0 g is allowed
The average rate of disappearance of ammonia gas was 166.39 torr*s-1
2NH3(g) -> N2(g) + 3H2(g) What is average rate of appearance of hydrogen?
the answer is:
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