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

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

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