JEE MAIN - Chemistry (2006 - No. 3)
The enthalpy changes for the following processes are listed below :
Cl2(g) = 2Cl(g), 242.3 kJ mol–1
I2(g) = 2I(g), 151.0 kJ mol–1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1
I2(s) = I2(g), 62.76 kJ mol–1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :
Cl2(g) = 2Cl(g), 242.3 kJ mol–1
I2(g) = 2I(g), 151.0 kJ mol–1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1
I2(s) = I2(g), 62.76 kJ mol–1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :
–14.6 kJ mol–1
–16.8 kJ mol–1
+16.8 kJ mol–1
+244.8 kJ mol–1
Explanation
$${{\rm{I}}_2}\left( s \right) + C{l_2}\left( g \right) \to 2{\rm{I}}Cl\left( g \right)$$
$$\Delta A = \left[ {\Delta {{\rm{I}}_2}\left( s \right) \to {l_2}\left( g \right) + \Delta {H_{{\rm I} - l}} + \Delta {H_{C{\rm I} - Cl}}} \right] - $$
$${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2\left[ {\Delta {H_{{\rm{I}} - Cl}}} \right]$$
$$ = 151.0 + 242.3 + 62.76 - 2 \times 211.3$$
$$ = 33.46$$
$$\Delta H_f^0\left( {{\rm{I}}Cl} \right) = {{33.46} \over 2}$$
$$ = 16.73\,\,kJ/mol$$
$$\Delta A = \left[ {\Delta {{\rm{I}}_2}\left( s \right) \to {l_2}\left( g \right) + \Delta {H_{{\rm I} - l}} + \Delta {H_{C{\rm I} - Cl}}} \right] - $$
$${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2\left[ {\Delta {H_{{\rm{I}} - Cl}}} \right]$$
$$ = 151.0 + 242.3 + 62.76 - 2 \times 211.3$$
$$ = 33.46$$
$$\Delta H_f^0\left( {{\rm{I}}Cl} \right) = {{33.46} \over 2}$$
$$ = 16.73\,\,kJ/mol$$
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