Question

the heat of combustion of naphthalene at constant volume at 25 degree celsius was found to be -5133 kj/mole calculate the value of enthalpy change at constant pressure at the same temperature

Answer 1

Answer : The value of enthalpy change at constant pressure at the same temperature is -5137.9 kJ

Explanation :

Heat released at constant pressure is known as enthalpy.

Heat released at constant volume is known as internal energy.

The balanced combustion reaction will be,

$C_{10}H_8(s)+12O_2(g)\rightarrow 10CO_2(g)+4H_2O(l)$

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$q_p=q_v+\Delta n_gRT$

where,

$q_p$ = heat change at constant pressure  = ?

$q_v$ = heat change at constant volume  = -5133 kJ/mole

$\Delta n_g$ = change in moles = [(10)-(12)] = -2 mole

R = gas constant  = $8.314\times 10^3kJ/mole.K$

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above formula, we get:

$q_p=q_v+\Delta n_gRT$

$q_p=(-5133kJ/mole)+(-2mole)\times (8.314\times 10^{-3}kJ/mole.K)\times (298K)$

$q_p=-5137.9kJ$

Therefore, the value of enthalpy change at constant pressure at the same temperature is -5137.9 kJ

Answer 2

Answer:

-5137 kJ/mol

Explanation:

qv=-5133kJ/mol

= -5133X1000 J/mol

T= 25C

= 25+273= 298K

∆n=10-12= -2 mol

qp= qv + ∆nRT

= -5133000-2X8.314X298

= - 5137,955.14

= -5137.9 kJ/mol

qp= -5137.9kJ/mol