Q12.137CP - Chemistry: Molecular Nature Of Matter And Change

Question

Substance A has the following properties. 

$$\begin{gathered} \mathbf{mp}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{1}\mathbf{atm}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{20}\mathbf{.}\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{bp}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{atm}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{85}\mathbf{°}\mathbf{C} \\ \mathbf{∆}{\mathbf{H}}_{\mathbf{fus}}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{180}\mathbf{.}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{∆}{\mathbf{H}}_{\mathbf{vap}}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{500}\mathbf{.}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g} \\ {\mathbf{c}}_{\mathbf{solid}}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{·}\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{c}}_{\mathbf{liquid}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{·}\mathbf{°}\mathbf{C} \\ {\mathbf{c}}_{\mathbf{gas}}\mathbf{:}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{·}\mathbf{°}\mathbf{C} \end{gathered}$$

                                           

At 1 atm, a 25-g sample of A is heated from  $\mathbf{-}\mathbf{40}\mathbf{.}\mathbf{°}\mathbf{C}$ to $\mathbf{100}\mathbf{.}\mathbf{°}\mathbf{C}$  at a constant rate of 450.0 J/min.

(a) How many minutes does it take to heat the sample to its melting point? 

(b) How many minutes does it take to melt the sample? 

(c) Perform any other necessary calculations, and draw a curve of temperature vs. time for the entire heating process.

Step-by-Step Solution

Answer

(a) The time taken by the heat to melt the sample is 1.1 minutes.

(b) The time taken to melt the sample is 6.7 minutes.

(c) The change in the phase from the solid to liquid then liquid to gaseous with the change in temperature.

Definition

Enthalpy of vaporisation may be defined as the as the energy required to convert the liquid into vapour phase. It is also known as Enthalpy of evaporation.

 

Clausius- clapeyron equation

 $\ln \frac{\mathrm{P}2}{\mathrm{P}1}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}\mathrm{vap}}{\mathrm{R}}\left[\frac{1}{\mathrm{T}2}-\frac{1}{\mathrm{T}1}\right]$

 

P = Pressure

T = Temperature

R = Gas Constant

$∆{\mathrm{H}}_{\mathrm{vap}}^{°}$ = Enthalpy of vaporisation

 

Ideal gas equation is a type of equation that defined the ideal gas. The ideal gas does not affect by the change in the pressure or volume of the gas at any temperature.

 

The equation for the ideal gas:

 $\mathbf{Heat}\mathbf{ }\mathbf{Energy}\mathbf{,}\mathbf{Q}\mathbf{=}\mathbf{m}\mathbf{\times }{\mathbf{C}}_{\mathbf{solid}}\mathbf{\times }\mathbf{\Delta T}$

 

${\mathrm{C}}_{\mathrm{solid}}$ = Specific heat of solid,  

M = Mass 

T = Temperature.

Numerical Explanation

(a) Change in Temperature = $-40{.}^{\mathrm{o}}\mathrm{C}$ to  $100{.}^{\mathrm{o}}\mathrm{C}$

The melting point of the sample is $-20{.}^{\mathrm{o}}\mathrm{C}$

Firstly, for reaching the sample is heated from $-40{.}^{\mathrm{o}}\mathrm{C}$ to $-20{.}^{\mathrm{o}}\mathrm{C}$ temperature.

Specific heat of the Solid,  $\text{Csolid}:1.0\frac{\text{J}}{{\text{g}}^{\text{o}}\text{C}}$

The heat energy required for raising the temperature from $-40°C$ to $-20°C$$\begin{array}{rcl}\mathrm{Heat} \mathrm{Energy}, \mathrm{Q}& =& \mathrm{m}\times {\mathrm{C}}_{\mathrm{solid}}\times ∆\mathrm{T}\\ & =& 25 \mathrm{g}\times 1.0 \mathrm{J}/\mathrm{g}°\mathrm{C}\times \left[\left(-20°\mathrm{C}-\left(-40°\mathrm{C}\right)\right)\right]\\ & =& 25 \mathrm{g}\times 1.0 \mathrm{J}/\mathrm{g}°\mathrm{C}\times 20°\mathrm{C}\\ & =& 500 \mathrm{J}\end{array}$

 

Rate constant of the equation = 450J/min

$$\begin{gathered} \text{Time}=\frac{\text{Heat} \text{Energy}}{\text{Rate} \text{Constant}} \\ \text{Time}=\frac{500\text{J}}{450\text{J}/\text{min}} \\ \text{Time}=1.1 \text{min} \end{gathered}$$

 

The time taken by the heat to melt the sample is 1.1 minutes.

 

(b) Change in Temperature = $-20°C$ to  $100°C$

The melting point of the sample is  $-20°C$

Firstly, for reaching the sample is heated from $-20°C$ to $100°C$ temperature.

Specific heat of the Solid,  $\text{Csolid:}1.0\frac{\text{J}}{\text{g}°\text{C}}$

The heat energy required for raising the temperature from $-20°C$ to $100°C$:

 $$\begin{gathered} \mathrm{Heat} \mathrm{Energy},\mathrm{Q}=\mathrm{m}\times {\mathrm{C}}_{\mathrm{solid}}\times \mathrm{\Delta T} \\ \mathrm{Q}=\left(25\mathrm{g}\right)\times 1.0\frac{\mathrm{J}}{{\mathrm{g}}^{\mathrm{o}}\mathrm{C}}\times \left[\left({100}^{\mathrm{o}}\mathrm{C}\right)-\left(-{20}^{\mathrm{o}}\mathrm{C}\right)\right] \\ \mathrm{Q}=\left(25\mathrm{g}\right)\times 1.0\mathrm{J}/{\mathrm{g}}^{\mathrm{o}}\mathrm{C}\times {120}^{\mathrm{o}}\mathrm{C} \\ \mathrm{Q}=3000\mathrm{J} \end{gathered}$$

 

Rate constant of the equation = 450J/min

$$\begin{gathered} \text{Time}=\frac{\text{Heat} \text{Energy}}{\text{Rate} \text{Constant}} \\ \text{Time}=\frac{3000\text{J}}{450\frac{\text{J}}{\text{min}}} \\ \text{Time}=6.7 \text{min} \end{gathered}$$

 

The time taken by the heat to melt the sample is 6.7 minutes.

Explanation for (c)

(c) The graph of temperature (y-axis) vs time (x-axis):

 

The change in the phase from the solid to liquid then liquid to gaseous with the change in temperature.