Question

For a particular isomer of CH 8, the combustion reaction produces 5093.7 kJ of heat per mole of C,H,8(g) consumed, under standard conditions. CH 800,(9) +9H 0(g) AH; n = -5093.7 kJ/mol What is the standard enthalpy of formation of this isomer of C,H,8(E)? AH;= kJ/mol wimol

Answer 1

We know that $\Delta$H^o_rxn = $\Delta$H^o_products - $\Delta$H^o_reactants ----------------------- (1)

It is given, $\Delta$H^o_rxn = -5093.7kJ/mol

$\Delta$H^o_products = [8x $\Delta$H^o_f (CO_2(g))] + [9x $\Delta$H^o_f (H₂O_(g))] -------------------- (2)

$\Delta$H^o_f (CO_2(g)) = -393.5kJ /mol

$\Delta$H^o_f (H₂O_(g)) = -241.8 kJ/mol

Substitute this values in equation (2), we get

$\Delta$H^o_products = 8 x (-393.5kJ /mol) + 9 x (-241.8 kJ/mol)

= -3148 kJ/mol +(- 2176.2 kJ/mol)

= - 5324.2 kJ/mol

$\Delta$H^o_reactants = $\Delta$H^o_f (C₈H_18(g)) + (25/2 x $\Delta$H^o_f (O_2(g)))

$\Delta$H^o_f (C₈H_18(g)) = ? let it be x

$\Delta$H^o_f (O_2(g)) = 0

$\Delta$H^o_reactants = x -0

= x

Substitute this values in equation (1), to get x value.

-5093.7kJ/mol = - 5324.2 kJ/mol - x

=> -x = -5093.7kJ/mol +  5324.2 kJ/mol

=> - x = 230.5 kJ/mol

=> x = -230.5 kJ/mol

hence,

$\Delta$H^o_f (C₈H_18(g)) = -230.5 kJ/mol