equilibrium constant (1 Viewer)

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2NO(g) + O2(g) ---> 2NO2(g)

a 1 litre reaction vessel initially contained 0.25 mol NO and 0.12 mol O2. After equilibrium was established there was only 0.05 mol NO.

calculate the equilibrium constant

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just need to check my answer for this question (i got 6.4 L/mol which looks abnormal).

thanks in advance
 

b3kh1t

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2NO(g) + O2(g) ---> 2NO2(g)

a 1 litre reaction vessel initially contained 0.25 mol NO and 0.12 mol O2. After equilibrium was established there was only 0.05 mol NO.

calculate the equilibrium constant

---

just need to check my answer for this question (i got 6.4 L/mol which looks abnormal).

thanks in advance
the molar ratios are; NO : O2 : NO2 = 2 : 1 : 2 moles
initially = 0.25 : 0.12 : 0 moles
moles used/formed = 0.20 : 0.10 : 0.20 moles
at equilibrium = 0.05 : 0.02 : 0.20 moles
therefore since the volume of the vessel is 1L, the concentrations are; [NO]=0.05 M, [O2]=0.02 M, [NO2]=0.20 M

the equilibrium constant is given by
therfore = 800 L/mol (for HSC the units is not required but that is correct)
 

xDarkSilent

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Remember this :

A + B --> C + D

equilibrium constant = [C][D] / [A]

2A + B --> 3C + D

Equilibrium constant = [C]^3[D] / [A]^2

The mole ratio's becomes the power of it in the equilibrium constant equation

^above user gave the correct answer.
 

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