3/ At 2000 K , nitrogen gas and oxygen gas combine to form nitric oxide accordin
ID: 882295 • Letter: 3
Question
3/ At 2000 K , nitrogen gas and oxygen gas combine to form nitric oxide according to the following reaction: N2(g) + O2(g) 2NO(g) H = +1.81 KJ
with an equilibrium constant K of 4.1 x 10^-4. If 0.50 mole of N2 and 0.86 mole of O2 are put into a 2.0 L container at 2000 K , what would the equilibrium concentrations of all species be ?
4/ For the reaction in question #3. How would the following changes effect the equilibrium ? Each change occurs independently of the others. Increase temperature. Decrease pressure. Decrease [O2(g)]. Increase [NO(g)]
Explanation / Answer
3) N2(g) + O2(g) 2NO(g)
Inital concentration 0.5 / 2 0.86 /2 0
Change -x -x 2x
Equilbrium concent 0.25-x 0.43-x 2x
Kc = (2x)2 / (0.25 -x) ( 0.43-x) = 4.1 X 10^-4
4.1 X 10^-4 = 4x2 / (0.1075 -0.25x - 0.43 x + x2)
we can ignore x2 in denominator as it is very less as Kc is very low.
0.00041 (0.1075 - 0.68x ) = 4x2
0.0000441 - 0.000278x = 4x2
4x2 + 0.000278x -0.0000441 = 0
on solving
x = 0.00328
So equilibrium concentration of each species will be
[N2] = 0.25 - 0.00328 = 0.246
[O2] = 0.43-0.00328 =0.426
[NO] = 2X0.00328 = 0.00656
4) a) the enthalpy of reaction is positive so reaction is endothermic in nature, on increasing temperature the reaction will go forward direction ( Le-Chatelier's priciple)
b) decrease pressure will not affect the equilibrium as number of moles of reactants = number of moles of products = 2
c) decreasing conncentration of O2 ( or N2) will shift the equilbrium toward backward direction
d) Increase in concentration of NO (product) will shift the equilbrium toward backward direction
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