Question

⏺️At 300K the standard enthalpies of formation of
C6H5COOH(s), CO2(g) and H20(l) are -408, -393
and -286 kJ mol^1 respectively. Calculate the heat
of combustion of benzoic acid at contant volume :
(1) +3201 kJ
(2) +3199.75 kJ
(3) -3201 kJ
(4) -3199.75 kJ

⏺️⏺️ Explain your answer in detail..⏺️⏺️​

Answer 1

Answer:

can you pls elaborate the question. ( without the emojis )

Answer 2

Answer :

We have been provided values of the standard enthalpies of formation$.$

• C₆H₅COOH = -408 kJ/mol
• CO₂ = -393 kJ/mol
• H₂O = -286 kJ/mol

We have to find the heat of combustion of benzoic acid.

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◈ Balanced chemical equation :

$\boxed{\bf{\purple{C_6H_5COOH+\dfrac{17}{2}O_2}\:\longrightarrow\:\orange{7CO_2+3H_2O}}}$

◈ Heat of combustion :

$\sf{\Delta_cH^{\circ}=\sum(\Delta_fH^{\circ})_{products}-\sum(\Delta_fH^{\circ})_{reactants}}$

$\sf{\Delta_cH^{\circ}=[7(\Delta_fH^{\circ}_{CO_2})+3(\Delta_fH^{\circ}_{H_2O})]-[\Delta_fH^{\circ}_{C_6H_5COOH}]}$

$\sf{\Delta_cH^{\circ}=[7(-393)+3(-286)]-(-408)}$

$\sf{\Delta_cH^{\circ}=(-2751-858)-(-408)}$

$\sf{\Delta_cH^{\circ}=-3609+408}$

$\bf{\Delta_cH^{\circ}=-3201\:kJ}$