Equality reaction

[Mathman23]

Hi

I got two questions:

1] A equality reaction:
\mathrm{2N + O_{2} \rightleftharpoons 2NO}

a) How does one remove NO from the reaction?


2] An equality mixture \mathrm{H_2 + Br_2 \rightleftharpoons 2 HBr}

contains 0,340 mol H_2 , 0,220 mol \mathrm{Br_2}.

in a canister which has a volume V = 1,0 L.

The mixture is heated to reaches a tempeture of 700K.

Then the reaction has set itself 0,0196 mol is left of Br_{2}

a) The concentration of \mathrm{C_{H_2} = \frac{0,34 mol}{1,0 L} = 0,34 mol/L}

The concentration of \mathrm{C_{Br_2} = \frac{0,22 mol}{1,0 L} = 0,22 mol/L}

b) The equality fraction for the reaction is: \mathrm{K_c} = \frac{\mathrm{[HBr]^2}}{\mathrm{[H_2] \cdot [Br_2]}}

c) How do I obtain [HBr] ?

Sincerely

/Fred

[chem_tr]

Hello,

An equilibrium may shift either side if the concentration of the contributors can be changed. You may try this by further reacting with O2 to yield NO2.

NO+\frac {1}{2} O_2 \leftrightharpoons NO_2

About your second question, I think you are finding the initial concentrations. This may be incorrect as you seem to omit something. At 700°C, the available hydrogen is completely reacted with oxygen to give water. I don't know any change would be present in HBr and Br2.

You may find that 0,120 mols of hydrogen is excess in the medium, so it can be converted to water via the reaction H_2+ \frac {1}{2}O_2 \rightarrow H_2O. After the equilibrium is reached, you state that 0,0196 moles of bromine is present. You can do the rest from here, if you're stuck somewhere, don't hesitate to ask.

[wolfson_1123]

Hi there,

\mathrm{K_c} = \frac{\mathrm{[HBr]^2}}{\mathrm{[H_2] \cdot [Br_2]}}

Where i = initial and f = final

[H2init--][Br2init]--[2HBr2init]
[0.340--][0.220 ]--[2---x---]
[H2Fin--]Br2Fin ]--[2HBr2fin]
[0.1396 ][0.0196]--[2---x--]

/_\ = 0.2004

So 2HBr = 2 x /_\ = 2 x 0.2004.

So Concentration = 2/_\ / Volume = 2 x 0.2004 / 1 = well you can do that :-)

And then you can easily calculate the Kc.

Wolfson.