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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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The heat of combustion of naphthalene $(C_{10}H_8(s))$ at constant volume was measured to be $-5133\: kJ/ mol$ at $298\: K$. Calculate the value of enthalpy change $(R = 8.314 \: J/Kmol)$

$\begin {array} {1 1} (a)\;5128.04\: kJ & \quad (b)\;-5128.04\: kJ \\ (c)\;513.7955\: kJ & \quad (d)\;-5137955.14 \: J \end {array}$

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The combustion equation of naphthalene is
$C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l); \: \Delta E = -5133 kJ$
$\Delta n = 10-12 = -2 mol$
Now, applying the relation, $ \Delta H = \Delta E + (\Delta n)RT$
$ = -5133\: 5128.04\: kJ \: 10^3J + (-2\: mol)(8.314 \: J/mol \: K)(298 \: K)$
$= -5133000 \: J – 4955.14\: J = -5137955.14 \: J$
Ans : (d)
answered Mar 13, 2014 by thanvigandhi_1
 

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