Question

in habers process 30 litres of dihydrogen and 30 litres of dihydrogen were taken for reaction which yielded only 50% of the expected product what will be the composition of gaseous mixture under the aforesaid condition in the end?

Answer 1

From the given,

$Volume\quad of\quad { N }_{ 2 }\quad =\quad 30\quad L$

$Volume\quad of\quad { H }_{ 2 }\quad =\quad 30\quad L$

In Haber process, the reaction between nitrogen and hydrogen is as follows ,

${ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad N }_{ 2 }\quad \quad +\quad 3{ H }_{ 2 }\quad \rightarrow \quad 2N{ H }_{ 3 }$

$Initial\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 30\quad L\quad \quad \quad 30\quad L\quad \quad \quad \quad$

$After\quad some\quad time\quad 10\quad \quad \quad \quad \quad \quad 30\quad \quad \quad \quad \frac { 2 }{ 3 } \times 30\\ Reaction\quad left\quad \quad \quad \quad 20\quad L\quad \quad \quad \quad 0\quad L\quad \quad \quad \quad 20\quad L$

$But,\quad from\quad the\quad given\quad N{ H }_{ 3 }\quad consume\quad 50%$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 20\quad \times \quad \frac { 50 }{ 100 } \quad =\quad 10\quad L\quad$

$\quad \quad \quad \quad \quad \quad \quad { N }_{ 2 }\quad \quad =\quad \frac { 10 }{ 2 } \quad =\quad 5\quad L$

$\quad \quad \quad \quad \quad \quad \quad left\quad =\quad 25\quad L$

$\quad \quad \quad \quad \quad \quad { H }_{ 2 }\quad left\quad =\quad 30\quad \quad -\quad \frac { 10 }{ 2 } \quad \times \quad 3\quad =\quad 15\quad L$