Question

How much energy (in kJ) is released when 26.3 g of NH3 is formed?

Answer 1

Energy released:

`\<strong>Delta H=-71.03 kJ</strong>`

Step-by-step explanation:

In Haber's process Ammonia formation takes place according to equation:

`<strong>N_(2)+3H_(2)\leftrightharpoons 2NH_(3)</strong>`

`\<strong>Delta H=-92 kJ/mol</strong>`

According to this equation ,when 2 mole of ammonia is formed from 3 mole of nitrogen and 1 mole of oxygen then 92 kJ of heat is released.

1 mole of
`<strong>NH_(3)</strong>` contains 17.03 g

mass of N=14.0067 g/mol

mass of H=1.007 g/mol

mass of
`NH_(3)` =14.0067+3.021

mass of
`<strong>NH_(3)</strong>` =17.03 g/mol

So, 2 mole of
`<strong>NH_(3)</strong>` contains
`17.03* 2`=34.06g

Since 34.06g (or 2mol) = -92 kJ

`1 g=(92)/(34.06)`

`<strong>26.3 g=(92)/(34.06)* 26.3</strong>`

this gives
`\<strong>Delta H=-71.03 kJ</strong>`

Hence amount of energy released when 26.3g of ammonia is formed is -71.03 kJ