Question

A solution is $500ml$ of $2MKOH$ is added to $500ml$ of $2MHCl$ and the mixture is well shaken. The rise in temperature $T_1$ is noted. The experiment is then repeated using $250ml$ of each solution and rise in temperature $T_2$ is again noted. Assume all heat is taken by the solution.

• A. $T_1=T_2$
• B. $T_1$ is $2$ times as large as $T_2$
• C. $T_2$ is twice of $T_1$
• D. $T_1$ is $4$ times as large as $T_2$

Answer 1

The correct option is A $T_1=T_2$

Let heat of reaction be $Q$

$KOH+HCl\to FCl+H_{2}O\Delta H=-Q$

$Q$ heat released for $1$ mole of salt

we know that

$Q=Ms\Delta T$

$M=Jv$ [$J:$ density, $V:$ volume]

$Q=Jv\Delta TS$

$Q\times v\times \Delta T$

$\Rightarrow \frac{Q_1}{Q_2}=\frac{V_1{T}_1}{V_2{T}_2}$

Here $Q_1=Q$ [$\because$ $1$4 mole salt formed]

Here $Q_2=Q/2$ ($\because$ $1/2$ mole salt formed)

$v_1=1000ml,v_2=500ml$

$\frac{\left(Q\right)}{\left(Q/2\right)}=\frac{1000}{500}\frac{T_1}{T_2}$

$\Rightarrow T_1=T_2$