Chapter EN, Problem 3PQ

Interpretation Introduction

Interpretation:

Relationship between $\Delta T_1$ and $\Delta T_2$ for solution of $0.5\text{M NaOH}$ with $0.5\text{M HCl}$ has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Answer to Problem 3PQ

Option (D) is the correct one.

Explanation of Solution

Reason for correct option:

The formula to determine molarity is as follows:

  $\text{molarity}=\frac{\text{number}\text{of}\text{moles}}{\text{volumeof solution}}$        (1)

Rearrange equation (1) to calculate number of moles of $\text{NaOH}$.

  $\text{number}\text{of}\text{moles}=\left(\text{molarity}\right)\left(\text{volume of solution}\right)$        (2)

Substitute $0.5\text{M}$ for molarity and $100\text{mL}$ for volume of solution in equation (2).

  $\begin{array}{c}\text{number}\text{of}\text{moles}=\left(0.5\text{M}\right)\left(100\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{\text{mL}}\right)\\ =0.05\text{mol}\end{array}$

Neutralization reaction between $\text{NaOH}$ and $\text{HCl}$ occurs as follows:

  $\text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_{\text{2}}\text{O}$

Since $1\text{M NaOH}$ react with $1\text{M HCl}$ to produce $1{\text{M H}}_{\text{2}}\text{O}$ therefore $0.05\text{M NaOH}$ will react with $0.05\text{M HCl}$ to produce $0.05{\text{M H}}_{\text{2}}\text{O}$.

Consider heat energy associated with neutralization reaction to produce $0.05{\text{M H}}_{\text{2}}\text{O}$ be $q$. This $q$ will heat $\text{200 mL}$ solution. Then mass of solution will be $200\text{g}$ as density of water is $1\text{g/mL}$.

Heat energy associated with neutralization reaction to produce $0.05{\text{M H}}_{\text{2}}\text{O}$ is as follows:

  $q=mC\left(\Delta {\text{T}}_1\right)$        (3)

$\Delta T_1$ is temperature difference associated with neutralization reaction to produce $0.05{\text{M H}}_{\text{2}}\text{O}$.

Rearrange equation (3) to calculate $\Delta T_1$.

  $\Delta {\text{T}}_1=\frac{mC}{q}$        (4)

Substitute $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$ for $C$ and $200\text{g}$ for $m$ in equation (4) to $\Delta T_1$.

  $\begin{array}{c}\Delta {\text{T}}_1=\frac{\left(200\text{g}\right)\left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)}{q}\\ =\frac{836.8\text{J}\cdot {\text{K}}^{-1}}{q}\end{array}$

Similarly, when $200\text{mL}$ of each $\text{NaOH}$ and $\text{HCl}$ is mixed then moles of water formed will have moles as follows:

  $\begin{array}{c}\text{number}\text{of}\text{moles}=\left(0.5\text{M}\right)\left(200\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{\text{mL}}\right)\\ =0.1\text{mol}\end{array}$

Since $0.1\text{mol}$ of ${\text{H}}_2\text{O}$ is twice of $0.05{\text{M H}}_{\text{2}}\text{O}$, therefore, heat energy associated to produce $0.1\text{mol}$ will $2q$.

Heat energy associated with neutralization reaction to produce $0.1{\text{M H}}_{\text{2}}\text{O}$ is as follows:

  $2q=mC\left(\Delta {\text{T}}_2\right)$        (5)

$\Delta T_2$ is temperature difference associated with neutralization reaction to produce $0.1{\text{M H}}_{\text{2}}\text{O}$.

Rearrange equation (5) to calculate $\Delta T_2$.

  $\Delta {\text{T}}_2=\frac{mC}{2q}$        (6)

$\text{200 mL}$ of $\text{NaOH}$ and $\text{HCl}$ makes $\text{400}\text{mL}$ water solution. The density of water is $1\text{g/mL}$ so mass of water is $400\text{g}$.

Substitute $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$ for $C$ and $400\text{g}$ for $m$ in equation (6) to $\Delta T_2$.

  $\begin{array}{c}\Delta {\text{T}}_2=\frac{\left(400\text{g}\right)\left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)}{2q}\\ =\frac{\left(200\text{g}\right)\left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)}{q}\\ =\frac{836.8\text{J}\cdot {\text{K}}^{-1}}{q}\end{array}$

Relationship between $\Delta T_1$ for $\text{100 mL}$ each of $0.5\text{M NaOH}$ with $0.5\text{M HCl}$ and $\Delta T_2$ for $\text{200 mL}$ of each is as follows:

  $\Delta T_1=\Delta T_2$

Hence, option (D) is correct.

Reason for incorrect option:

Relationship between $\Delta T_1$ for $\text{100 mL}$ each of $0.5\text{M NaOH}$ with $0.5\text{M HCl}$ and $\Delta T_2$ for $\text{200 mL}$ is as follows:

  $\Delta T_1=\Delta T_2$

Therefore, choice (A),(B) and (C) is incorrect.

Conclusion

Relationship between $\Delta T_1$ for $\text{100 mL}$ each of $0.5\text{M NaOH}$ with $0.5\text{M HCl}$ and $\Delta T_2$ for $\text{200 mL}$ of each is as follows:

  $\Delta T_1=\Delta T_2$

Hence option (D) is correct.