Heat capacity $\left(C_p\right)$ is an extensive property but specific heat $(c)$ is intensive property. What will be the relation between $C_p$ and $c$ for 1 mole of water?
For water, molar heat capacity,
$$\begin{aligned} C_p & =18 \times \text { Specific heat, } \mathrm{c} \\ C_p & =18 \times c \text { Specific heat } \\ c & =4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} \quad \text{(for water)} \end{aligned}$$
Heat capacity,
$$\begin{aligned} C_p & =18 \times 4.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & =75.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}$$
The difference between $C_p$ and $C_v$ can be derived using the empirical relation $H=U+p V$. Calculate the difference between $C_p$ and $C_v$ for 10 moles of an ideal gas.
Given that, $C_v=$ heat capacity at constant volume,
$C_p=$ heat capacity at constant pressure
Difference between $C_p$ and $C_v$ is equal to gas constant ($R$).
$$\begin{aligned} \therefore \quad C_p-C_v & =n R \quad \text{(where, $n=$ no. of moles)}\\ & =10 \times 4.184 \mathrm{~J} \\ & =41.84 \mathrm{~J} \end{aligned}$$
If the combustion of 1 g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Given that, enthalpy of combustion of 1 g graphite $=20.7 \mathrm{~kJ}$
Molar enthalpy change for the combustion of graphite, $\Delta H=$ enthalpy of combustion of 1 g graphite $\times$ molar mass
$\begin{aligned} & \Delta H=-20.7 \mathrm{kJg}^{-1} \times 12 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \Delta H=-2.48 \times 10^2 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
Negative sign in the value of $\Delta H$ indicates that the reaction is exothermic.
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction? $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HBr}(\mathrm{g})$. Given that, bond energy of $\mathrm{H}_2, \mathrm{Br}_2$ and HBr is 435 $\mathrm{kJ} \mathrm{mol}^{-1}$, $192 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $368 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
Given that, bond energy of $\mathrm{H}_2=435 \mathrm{~kJ} \mathrm{~mol}^{-1}$
bond energy of $\mathrm{Br}_2=192 \mathrm{~kJ} \mathrm{~mol}^{-1}$
bond energy of $\mathrm{HBr}=368 \mathrm{~kJ} \mathrm{~mol}^{-1}$
For the reaction
$$\begin{aligned} &\begin{aligned} & \mathrm{H}_2(g)+\mathrm{Br}_2(g) \rightarrow 2 \mathrm{HBr}(g) \\ & \Delta_r H^s=\Sigma \mathrm{BE} \text { (Reactants) }-\Sigma \mathrm{BE} \text { (Products) } \\ & =\mathrm{BE}\left(\mathrm{H}_2\right)+\mathrm{BE}\left(\mathrm{Br}_2\right)-2 \mathrm{BE}(\mathrm{HBr}) \\ & =435+192-(2 \times 368) \mathrm{kJ} \mathrm{mol}^{-1} \\ & =-109 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\\ \end{aligned}$$
The enthalpy of vaporisation of $\mathrm{CCl}_4$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the heat required for the vaporisation of 284 g of $\mathrm{CCl}_4$ at constant pressure. (Molar mass of $\mathrm{CCl}_4=154 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$$\begin{aligned} & \text { Given that, } 1 \mathrm{~mol} \mathrm{of} \mathrm{CCl}_4=154 \mathrm{~g} \\ & \Delta_{\text {vap }} H \text { for } 154 \mathrm{~g} \mathrm{CCl}_4=30.5 \mathrm{~kJ} \\ & \therefore \quad \Delta_{\text {vap }} H \text { for } 284 \mathrm{~g} \mathrm{CCl}_4=\frac{30.5 \times 284}{154} \mathrm{~kJ}=56.25 \mathrm{~kJ} \end{aligned}$$