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Please refer to the given set of reactions with the respective enthalpy change

$\small i.\: H(g)+Cl(g)\rightarrow HCl(g)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \Delta H_1=-430kJ$

$\small ii.\: HCl(g)+aq\rightarrow H^\oplus(aq) +Cl^\ominus (aq)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \Delta H_2=-80kJ$

$\small iii.\: H(g)+aq\rightarrow H^\oplus(aq) +e^-\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \Delta H_3=1320kJ$

$\small iv.Cl(g)+e^-\rightarrow Cl^\ominus(g) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \Delta H_4=-350kJ$

Calculate the enthalpy of hydration of $\small Cl^\ominus$ ions if the enthalpy of hydration of $\small H^\oplus$ is zero

$\small H^\oplus(g)+Cl^\ominus (g)+aq\rightarrow H^\oplus (aq)+Cl^\ominus (aq)$
Option: 1
+1480 KJ

Option: 2
-1480 KJ

Option: 3
-2180 KJ

Option: 4
+2180 KJ

Answers (1)

best_answer

If we add reactions i and ii and subtract iii and iv from it, then we will get the resultant reaction which is same as hydration enthalpy of HCl.

$\dpi{100} \small H^\oplus(g)+Cl^\ominus (g)+aq\rightarrow H^\oplus (aq)+Cl^\ominus (aq)$ ......(A)

$\Delta H=\Delta H_1+\Delta H_2-\Delta H_3-\Delta H_4$

$\Delta H=-430-80-1320-(-350)$

$\Delta H=-1480kJ$

Now we have given that enthalpy of hydration of $H^\oplus$ is zero which means,

$H^\oplus(g)+aq\rightarrow H^\oplus(aq),\: \: \: \: \Delta H=0$

Also the reaction (A) can be broken in two parts as

$Cl^\ominus(g)+aq\rightarrow Cl^\ominus(aq)$ ........... (1)

$H^\oplus(g)+aq\rightarrow H^\oplus(aq)$ ..............(2)

As the enthalpy change for reaction (2) is zero hence for reaction (1) the enthalpy change will be equal to $\Delta H$ of reaction A.

So for $Cl^\ominus(g)+aq\rightarrow Cl^\ominus(aq)\: \: \: \: \Delta H_2=\Delta H_A=-1480KJ$

Hence, option number (2) is correct

Posted by

Divya Prakash Singh

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