Question

Can someone help me? It needs to have a diagram that has arrows.

Answer 1

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Step-by-step explanation:

The balanced chemical reaction is,

`C_4H_(10)(g)+(13)/(2)O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)`

The expression for enthalpy change is,

`\Delta H=[n* H_f{products}]-[n* H_f{reactants}]`

Putting the values we get :

`\Delta H=[4* H_f_(CO_2)+5* H_f_(H_2O)]-[1* H_f_{C_4H_(10)}+(13)/(2)* H_f_(O_2)]`

`-2877=[(4* -393)+(5* -286)]-[1* H_f_{C_4H_(10)}+(13)/(2)* 0]`

`H_f_{C_4H_(10)=-125kJ/mol`

Thus enthalpy change for formation of butane is -125 kJ/mol