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28 gms of N₂ and 6 gms of H₂ were heated in a closed 1litre vessel. At equilibrium, 25.5 gms of NH₃ is present. The approximate value of K_C is

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3.16

21.33

5.55

2.54

answer is B.

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Detailed Solution

V_vessel = 1 lit

\large {\left( {{W_{{N_2}}}} \right)_i} = 28g

\large {\left( {{n_{{N_2}}}} \right)_i} = \frac{{28}}{{28}} = 1

\large \left( {{W_{{H_2}}}} \right) = 6g

\large {\left( {{n_{{H_2}}}} \right)_i} = \frac{6}{2} = 3

\large \left( {{W_{N{H_3}}}} \right)eq = 25.5g

\large {\left( {{n_{N{H_3}}}} \right)_{eq}} = \frac{{25.5}}{{17}} = 1.5

\large {N_2} + 3{H_2} \rightleftharpoons 2N{H_3}

1 mole of N₂ gives 2 moles of NH₃

'X₁' mole of N₂ gives 1.5 moles of NH₃

Number of moles of N₂ reacted = 0.75

3 moles of H₂ gives 2 moles of NH₃

'X₂' moles of H₂ gives 1.5 moles of NH₃

Number of moles of H₂ reacted = 2.25

\large \mathop {{N_2}}\limits^{1\,mole} \left( g \right) +

\large \mathop {3{H_2}}\limits^{3\,moles} \left( g \right)

\large {K_C} = \frac{{\left[ {N{H_3}} \right]}^2}{{\left[ {{N_2}} \right]}{\left[ {{H_2}} \right]^3}}

\large {K_C} = \frac{\left ( \frac{3}{2} \right )^2}{\left ( \frac{1}{4} \right )\left ( \frac{3}{4} \right )^3}

\large {K_C} = \frac{3}{2} \times \frac{3}{2} \times \frac{4}{1} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3}

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