A mixture of 3.32 moles of N2 and 3.32 moles of O2 is placed in a 1.00 L container at 448 °C and allowed to re?
A mixture of 3.32 moles of N2 and 3.32 moles of O2 is placed in a 1.00 L container at 448 °C and allowed to reach equilibrium according to the following equation.
N2(g) + O2(g) ⇌ 2NO(g)
The equilibrium constant is 100. Calculate the concentrations of N2 present at equilibrium.
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Hi Lauren,
K = p(NO) ² / ( p(N2) x p(O2) ) = 100 . . .
a million mole of something, to that end gases, includes 6.022 x 10^23 atoms if a million mole of each of those gases is placed in separate packing containers of 1L, then confident, all the gases could exhibit an identical stress. ideally, a million mole of gasoline additionally incorporate 22.4L of gasoline. while utilising the appropriate gasoline regulation, we see that pv = nrt if we don't understand the stress of the gasoline, we resolve for it p = nrt/v p = 1mole x 0.0821L-atm/mole-ok x 273K/1L and we do this for each gasoline, then all the gases that have an identical style of moles and quantity could have an identical stress p could = 22.4atm for all