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  2. N_{2} + 3H_{2}\rightarrow2NH_{3} After completion of 75% of the reaction what would be the mole fraction for the constituents
# Engineering

N_{2} + 3H_{2}\rightarrow2NH_{3} After completion of 75% of the reaction what would be the mole fraction for the constituents

Answers (1)
R Rakhi

As we learnt in

Mole Fraction -

It is the ratio of moles of solute or moles of solvent to moles of the solution.

- wherein

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n_{A} and n_{B} respectively; then the mole fractions of A and B are given as

Mole fraction of A = (number of moles of A)/(number of moles of solution ) = n_{A}/(n_{A}+ n_{B})

 

 At   t = \frac{3}{4},

                         N_{2} + 3H_{2}\rightarrow2NH_{3}

t = \frac{3}{4}               \frac{1}{4}            \frac{3}{4}                \frac{3}{2}

Moles of N2 = 0.25

Moles of H2 = 0.75

Moles of NH3 = 1.5

Mole fraction of  N_{2}=\frac{0.25}{2.5} =0.1

Mole fraction of  H_{2}=\frac{0.75}{2.5} =0.3

Mole fraction of  NH_{3}=\frac{1.5}{2.5} =0.6

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