Question

A mixture of 1. 57 mol of N2, 1. 92 mol of $H_{2}$ and 8.13 mol of $NH_{3}$ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction $N_{2}(g)+3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$ is 1.7 × $10^{2}$. Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?

Answer 1

The direction of the net chemical reaction depends upon the reaction constant ${ Q }_{ c }$ and equilibrium constant ${ K }_{ c }$.

Hence,

If the ${ Q }_{ c }\quad =\quad { K }_{ c }\quad \xrightarrow [ \quad  ]{ \quad  }$ reaction is at equilibrium state.

If the ${ Q }_{ c }\quad <\quad { K }_{ c }\quad \xrightarrow [ \quad  ]{ \quad  }$ reaction will take a forward direction.

If the ${ Q }_{ c }\quad >\quad { K }_{ c }\quad \xrightarrow [ \quad  ]{ \quad  }$ reaction will take a backward direction.

In the above case,

${ N }_{ 2(g) }\quad +\quad { 3H }_{ 2(g) }\quad \rightleftharpoons \quad { NH }_{ 3(g) } { Q }_{ c }\quad =\quad \frac { \left[ { N }_{ 2 } \right] \left[ { 3H }_{ 2 } \right]  }{ \left[ { 2NH }_{ 3 } \right]  }{ K }_{ c }\quad =\quad 1.7\quad \times \quad { 10 }^{ 2 }\quad or\quad 170$

Molecular Concentration (M) of reactants and products:

$\frac { 1.57\quad mol\quad of\quad { N }_{ 2 } }{ 20\quad L } \quad =\quad 0.078\quad M\quad { N }_{ 2 }\\ \frac { 1.58\quad mol\quad of\quad { H }_{ 2 } }{ 20\quad L } \quad =\quad 0.096\quad M\quad { H }_{ 2 }\\ \frac { 8.13\quad mol\quad of\quad { NH }_{ 3 } }{ 20\quad L } \quad =\quad 0.406\quad M\quad { NH }_{ 3 }$

Substituting the values:

${ Q }_{ c }\quad =\quad \frac { \left( 0.078 \right) \left( 0.096 \right)  }{ \left( 0.406 \right)  } \quad =\quad 0.018$

Comparing ${ K }_{ c } and { Q }_{ c }: 170\quad { K }_{ c }\quad >\quad 0.018$

Since the reaction constant ${ Q }_{ c }$ is < than equilibrium constant  ${ K }_{ c }$, the reaction will move in the forward direction and the reactants get converted into the products.