Question

15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Write the balanced equation and calculate the enthalpy change for this reaction.

Answer 1

Answer:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

$Entalpy=-2861.9~KJ$

Explanation:

In this case, we have to start with the reagents:

$Al~+~NH_4NO_3$

The compounds given by the problem are:

-) Nitrogen gas =  $N_2$

-) Water vapor  =  $H_2O$

-) Aluminum oxide =  $Al_2O_3$

Now, we can put the products in the reaction:

$Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_)$

When we balance the reaction we will obtain:

$2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)$

Now, for the enthalpy change, we have to find the standard enthalpy values:

$Al_(_S_)=0~KJ/mol$

$NH_4NO_3_(_a_q_)=-132.0~KJ/mol$

$N_2_(_g_)=0~KJ/mol$

$H_2O_(_g_)=~-~241.8~KJ/mol$

$Al_2O_3_(_S_)=~-~1675.7~KJ/mol$

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

$(0*2)~+~(-132*3)=~-396~KJ$

And the products:

$(0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ$

Finally, for the total enthalpy we have to subtract products by reagents :

$(-3125.9~KJ)-(-396~KJ)=-2729.9~KJ$

I hope it helps!