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 $\mathrm{CCl}_4=154 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$$\begin{aligned} & \text { Given that, } 1 \mathrm{~mol} \mathrm{of} \mathrm{CCl}_4=154 \mathrm{~g} \\ & \Delta_{\text {vap }} H \text { for } 154 \mathrm{~g} \mathrm{CCl}_4=30.5 \mathrm{~kJ} \\ & \therefore \quad \Delta_{\text {vap }} H \text { for } 284 \mathrm{~g} \mathrm{CCl}_4=\frac{30.5 \times 284}{154} \mathrm{~kJ}=56.25 \mathrm{~kJ} \end{aligned}$$
The enthalpy of reaction for the reaction
$$2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) \text { is } \Delta_r H^{\mathrm{s}}=-572 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
What will be standard enthalpy of formation of $\mathrm{H}_2 \mathrm{O}(l)$ ?
Given that,
$2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l), \Delta_r H^{\circ}=-572 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation is the enthalpy change of the reaction when 1 mole of the compound is formed from its elements then
$$\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta_r H^{\circ}=?$$
This can be obtained by dividing the given equation by 2.
Therefore, $$\Delta_f H^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-\frac{572 \mathrm{kJmol}^{-1}}{2}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{\text {ext }}$ in a single step as shown in figure? Explain graphically.
Suppose total volume of the gas is $V_i$ and pressure of the gas inside cylinder is $p$. After compression by constant external pressure, $\left(p_{\text {ext }}\right)$ in a single step, final volume of the gas becomes $V_f$.
Then volume change, $\Delta v=\left(V_f-V_i\right)$
If $W$ is the work done on the system by movement of the piston, then
$$\begin{aligned} & W=p_{\text {ext }}(-\Delta V) \\ & W=-p_{\text {ext }}\left(V_f-V_i\right) \end{aligned}$$
This can be calculated from $p-V$ graph as shown in the figure. Work done is equal to the shaded area $A B V_f V_i$
The negative sign in this expression is required to obtain conventional sign for $W$ which will be positive. Because in case of compression work is done on the system, so $\Delta V$ will be negative.
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
When compression is carried out in infinite steps with change in pressure, it is a reversible process. The work done can be calculated from $p-V$ plot as shown in the given figure. Shaded area under the curve represents the work done on the gas.
