Question

A solution of 200mL of 1M KOH is added to 200mL of 1M HCl and the mixture is well shaken. The rise in temperature ${T_1}$ is noted. The experiment is repeated by using ${\bf{100}}{{ }}{\bf{mL}}$ of each solution and increase in temperature ${T_2}$ is again noted. Which of the following is correct?
A.${T_1} = {T_2}$
B.${T_2}$ is twice as large as ${T_1}$
C.${T_1}$ is twice as large as ${T_2}$
D.${T_1}$ is four times as large as ${T_2}$

Answer 1

We know that the enthalpy of neutralisation for a strong acid and strong base is $- 57.1KJ$ . This is the energy that is released to the breaking of bonds in the acid and base. In this question, the second process considers half of the volume of the first process. Therefore the heat of neutralisation becomes half.
FORMULA USED: $M = \dfrac{n}{V}$
Where $M$ is the molarity, $n$ is the number of moles, $V$ is the volume.
Enthalpy of neutralisation = $Q = mC\Delta T$
Where $Q$ is the heat, $m$ is the mass, $C$ is the specific heat, $\Delta T$ is the difference in temperature.

Complete step by step answer:
As mentioned before, the reaction mentioned in this process is a neutralisation reaction that can be demonstrated below:
$KOH + HCl \to KCl + {H_2}O$
The volume of the ${\bf{1}}{{ }}{\bf{M}}{{ }}{\bf{KOH}}$ is ${\bf{200}}{{ }}{\bf{mL}}$ .
Therefore, the number of moles formed can be found used to the formula,
$M = \dfrac{n}{V}$
Plugging the values for molarity and volume, we get,
$1 = \dfrac{n}{{200 \times {{10}^{ - 3}}}}$
The volume is required to be obtained in litres. Therefore, it should be divided by $1000$ .
The number of moles of $KOH$ is
$n = 200 \times {10^{ - 3}}$
$n = 0.2$
Now we can find the number of moles of $HCl$ in the same way. This is carried out in the ssteps below:
$M = \dfrac{n}{V}$
Where $M$ is the molarity, $n$ is the number of moles, $V$ is the volume.
$1 = \dfrac{n}{{200 \times {{10}^{ - 3}}}}$
$n = 0.2$
For these reactions, the solution is treated as pure water. Therefore, we have to consider the volume of the solution is equal to the mass of the solution. Therefore, the total mass of the solution is,
${V_{TOTAL}} = {V_{KOH}} + {V_{HCl}}$
$\Rightarrow 200 + 200$
$\Rightarrow 400mL$
Therefore, the total mass of the solution is,
${m_{total}} = 400gms$
We also know that the heat of neutralisation can be found using the formula,
$Q = mC\Delta T$
In this case, the temperature change is ${T_1}$ .
Therefore, ${T_1}$ can be represented as follows,
$Q = 400 \times C{T_1}$
$\dfrac{Q}{{400 \times C}} = {T_1}$
In the second process the volume is halved, thus, the mass will be,
${V_{TOTAL}} = {V_{KOH}} + {V_{HCl}}$
$m = 100 + 100$
$\Rightarrow 200gms$
The number of moles in the second case is,
$n = \dfrac{{1 \times 100}}{{1000}}$
${n_{KOH}} = 0.1$
The number of moles of $KOH$ is also equal to the number of moles of $HCl$ .
Since the second reaction contains half the number of moles, the enthalpy of neutralisation will also be half.
Therefore, the relation will be as follows,
$\dfrac{Q}{2} = 200C{T_2}$
Taking $200$ to the denominator we get,
$\dfrac{Q}{{400C}} = {T_2}$ which is the same as the first difference in temperature.
Therefore, we can conclude by saying that, ${T_1} = {T_2}$ .
That is the answer to the question will be the option A.

Note: It is important to remember that in reactions of neutralisation the volume of a solution can be considered to be the mass of the solution as well. It is interchangeable. That is, if we look at the solution part, we can see that we take the volume of the solution to be the mass of the solution.
The heat of any reaction changes depending on the number of moles. Here, since the moles are halved, the heat of the reaction will also be halved.