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Yuliya22 [10]
3 years ago
5

15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Writ

e the balanced equation and calculate the enthalpy change for this reaction.
Chemistry
1 answer:
Lunna [17]3 years ago
8 0

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 <u>reagents</u>:

Al~+~NH_4NO_3

The compounds given by the problem are:

-) <u>Nitrogen gas</u> =  N_2

-) <u>Water vapor</u>  =  H_2O

-) <u>Aluminum oxide</u> =  Al_2O_3

Now, we can put the products in the <u>reaction</u>:

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

When we <u>balance</u> 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 <u>standard enthalpy values</u>:

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 <u>multiply</u> the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the <u>reagents</u>:

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

And the <u>products</u>:

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

Finally, for the total enthalpy we have to <u>subtract</u> products by reagents :

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

I hope it helps!

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