The enthalpy changes states for the following processes are listed $$belowCl2g$$ $$=$$ $$2Cl(g)$$; $$242.3$$ kJ $$mol-1$$ $$I2g$$ $$=$$ $$2I(g)$$; $$151.0$$ kJ $$mol-1$$ IClg $$=$$ $$I(g)+Cl(g)$$; $$211.3$$ kJ $$mol-1$$ $$I2s$$ $$=$$ $$I2(g)$$; $$62.76$$ kJ $$mol-1$$ Given that the standard states for iodine chlorine are $$I2s$$ and $$Cl2g$$, the standard enthalpy of formation for ICl (g) is

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The enthalpy changes states for the following processes are listed $$belowCl2g$$ $$=$$ $$2Cl(g)$$;                     $$242.3$$ kJ $$mol-1$$ $$I2g$$ $$=$$ $$2I(g)$$;                           $$151.0$$ kJ $$mol-1$$ IClg $$=$$ $$I(g)+Cl(g)$$;               $$211.3$$ kJ $$mol-1$$ $$I2s$$ $$=$$ $$I2(g)$$;                             $$62.76$$ kJ $$mol-1$$ Given that the standard states for iodine chlorine are $$I2s$$  and  $$Cl2g$$, the standard enthalpy of formation for ICl (g) is

Infinity Learn · Accepted Answer

$$I2+Cl2$$→$$2ICl$$ ∆$$H=eI2S$$→g $$+$$ $$eI-I$$ $$+eCl-Cl$$ $$-2$$ × $$eI-Cl$$  $$=62.76+151.0+242.3$$ $$-$$ $$2$$ × $$211.3$$ $$=33.46$$ kJ for $$2$$ moles of ICl ∴  ∆$$H/mol$$ $$=$$ $$33.462=16.73$$ kJ $$mol-1$$

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