Detemination of an Equilibrium Constant 1) 5.0 moles of NH3 and 5.0 moles of H2

Question

Detemination of an Equilibrium Constant

1) 5.0 moles of NH3 and 5.0 moles of H2 are introdued into a 2.00 L flask. At equilibrium 1.0 moles of NH3 gas remail. Calculate Kc for this reaction.

3H2(g) + N2(g) = 2NH3(g)

2) The Kc value for the following equilibrium at 500 degrees Celcius is 49.

H2(g) + I2(g) = 2HI(g)

If 1.00 moles of H2 and 2.00 moles of I2 are introduces into a 3.00 L flask at 500 degrees celcius, how many moles of HI are present at equilibrium?

3) What would happen to the values you calculated for Kc in the experiment if the Beer's Law constant (k) value you used was low (e.g. you used 6000 M^-1 when the actual constant had a value of 10,000 M^-1) ? Explain your answer.

4) 1/2N2(g) + O2(g) = NO2(g) Kc = 1.04 x 10^-9

N2(g) + 2O2(g) = N2O4(g) Kc = 7.42 x 10^-18

Calculate Kc for each of the following.

(a) 2NO2(g) = N2(g) +2O2(g)

(b) N2(g) + 2O2(g) = N2O2(g)

(c) 2NO2(g) = N2O4(g)

Explanation / Answer

1.

Reaction equation
3 H(g) + N(g) 2 NH(g)

Since 1mol of the initial 5 moles of ammonia remain. That means 4 moles were consumed as equilibrium has established. According to reaction equation 3 moles of hydrogen ar formed per 2 moles of ammonia consumed. i.e. the amount of hydrogen formed due to reaction is:

n(H) = (3/2)( - n(NH) ) = (3/2)( 4 mol ) = 6 mol

Furthermore one mole of nitrogen is formed per 2 moles of ammonia consumed:
n(N) = (1/2)( - n(NH) ) = (1/2)( 4 mol ) = 2 mol
So the amount in the equilibrium mixture are
n(NH) = 1 mol
n(H) = 5 mol + 6 mol = 11 mol
n(N) = 0 mol + 2 mol = 2 mol
The concentrations in the equilibrium mixture are:
[NH] = 1 mol / 2.0 L = 0.5 M
[H] = 11 mol / 2.0 L = 5.5 M
[H] = 2 mol / 2.0 L = 1 M
Hence,
Kc = 0.5² / ( 5.5³ 1 ) = 1.50×10³

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