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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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The enthalpy changes for the following processes are listed below \[\] $Cl_2(g) = 2Cl(g),\: \: \: \: \: 242.3\: kJ/mol$ \[\] $I_2(g) = 2I(g), \: \: \: \: \: 151\: kJ/mol$ \[\] $ ICl(g) = I(g) + Cl(g),\: \: \: 211.3 \: kJ/mol$ \[\] $ I_2(s) = I_2(g),\: \: \: \: 62.76\: kJ/mol$ \[\] Given that the standard states for iodine and chlorine are $I_2(g)\: and \: Cl_2(g)$, the standard enthalpy of formation of $ICl(g)$ is

$\begin {array} {1 1} (A)\;-14.6\: kJ/mol & \quad (B)\;-244.8\: kJ/mol \\ (C)\;16.8\: kJ/mol & \quad (D)\;244.8\: kJ/mol \end {array}$

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$ \large\frac{1}{2}$$ I_2(s) + \large\frac{1}{2}$$ Cl_2(s) \rightarrow ICl(g)$
$ \Delta H = [\large\frac{1}{2} $$ \Delta H_{s \rightarrow g} + \large\frac{1}{2} $$ \Delta H_{diss}. (Cl_2) + \large\frac{1}{2} $$ \Delta H_{diss}. (I_2)] - \Delta H_{ICl}$
$= ( \large\frac{1}{2} $$ \times 62.76 + \large\frac{1}{2}$$ \times 242.3 + \large\frac{1}{2}$$ \times151) – 211.3$
$= 228.03 – 211.3$
Thus, $ \Delta H =16.73$
Ans : (C)
answered Mar 18, 2014 by thanvigandhi_1
 

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