3 Marks Question

Question 513 Marks

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35^{\circ} \mathrm{C}$ to $55^{\circ} \mathrm{C}$. Molar heat capacity of Al is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.

Answer

Mass of Al = 60g Rise in temperature,$ \Delta\text{T} = 55 – 35 = 20°\text{C}$ Molar heat capacity of Al $= 24\text{J mol}^{-1} \text{K}^{-1}$ Specific heat capacity of Al $=\frac{24}{27}\text{Jg}^{-1} \ \text{K}^{-1}$$\therefore$ Energy required $ \text{m} \times \text{c} \times \Delta{\text{T}}$
= $60\times\frac{24}{27}\times20=\frac{28800}{27}=1066.67 \text{J}$
$= 1.068\text{kJ or 1.07kJ}$

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Question 523 Marks

The combustion of one mole of benzene takes place at 298K and 1atm. After combustion, $\mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ are produced and 3267.0kJ of heat is liberated. Calculate the standard enthalpy of formation of benzene $(\Delta_\text{f}\text{H}^\circ).$$\Delta_\text{f}\text{H}^\circ\text{CO}_2(\text{g})=-393\text{kJ mol}^{-1}$ and $\Delta_\text{f}\text{H}^\circ\text{H}_2\text{O(l})=-285.83\text{kJ mol}^{-1}.$

Answer

Combustion of 1 mole of benzene takes place as follows:$\text{C}_6\text{H}_6(\text{l})+\frac{15}{2}\text{O}_2\text{(g)}\overrightarrow{\ \ \ \ \ \ \ }\ 6\text{CO}_2\text{(g)}+3\text{H}_2\text{O(l)}$
$\Delta_\text{c}\text{H}^\circ=-3267.0\text{kJ mol}^{-1}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{r}\text{H}^\circ$
$=\sum[\Delta_\text{f}\text{H}^\circ(\text{products})]-\sum[\Delta_\text{f}\text{H}^\circ(\text{reactants})]$
$\text{or} -3267.0\text{kJ mol}^{-1}$
$ -3267.0\text{kJ mol}^{-1}=6[\Delta_\text{f}\text{H}^\circ\text{CO}_2(\text{g})]\\+3[\Delta_\text{f}\text{H}^\circ\text{H}_2\text{O(l)}]-1[\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})]\\-\frac{15}{2}[\Delta_\text{f}\text{H}^\circ\text{C}_2(\text{g})]$
$ -3267.0\text{kJ}=(6)(-393.5\text{kJ mol}^{-1})\\+(3)(-285.83\text{kJ mol}^{-1})-(1)[\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})]-0$
$\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})=[6(-393.5)\\+3(-285.83)+3267]\text{kJ mol}^{-1}$
$=[-2361.0-857.49+3267.0]\text{kJ mol}^{-1}$
$=48.51\text{kJ mol}^{-1}$

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Question 533 Marks

Two moles of an ideal gas initially at 27°C and one atmospheric pressure are compressed isothermally and reversibly till the final pressure of the gas is 10atm. Calculate q, W and $\Delta\text{U}$ for the process.

Answer

Here, $\mathrm{n}=2$ moles, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}, \mathrm{P}_1=1 \mathrm{~atm}, \mathrm{P}_2=10 \mathrm{~atm}$
$\text{W}=2.303\text{nRT}\log\frac{\text{p}_2}{\text{p}_1}$
$=2.303\times2\times8.314\text{JK}^{-1}\text{mol}^{-1}\times300\text{K}\\\times\log\frac{10}{1}=11488.28\text{J}$
For isothermal compression of ideal gas, $\Delta\text{U}=0$ Further, $\Delta\text{U}=\text{q}+\text{W}$$\therefore\text{q}=-\text{W}=-11488\text{J}$

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Question 543 Marks

The heat of combustion of $\mathrm{C}_2 \mathrm{H}_6$ is -368.4 kcal . Calculate heat of combustion of $\mathrm{C}_2 \mathrm{H}_4$, heat of combustion of $\mathrm{H}_2$ is $68.32 \mathrm{kcal} \mathrm{mol}{ }^{-1} . \Delta \mathrm{H}$ for the following reaction is $-37.1 \mathrm{kcal} . \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$

Answer

$\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{C}_2\text{H}_6(\text{g});$$\Delta\text{H}=-37.1\text{kcal}$
$\Delta_\text{c}\text{H}^{\ominus}\text{C}_2\text{H}_6=-368.4\text{kcal},$ $\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=?$
$\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})=-68.32\text{kcal}$
$\Delta\text{H}=\sum\Delta_\text{c}\text{H}^\ominus(\text{reactants})-\sum\Delta_\text{c}\text{H}^\ominus(\text{products})$
$\Delta\text{H}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4+\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})\\-\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_6(\text{g})$
$-37.1\text{kcal}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4-68.32-(-368.4)$
$\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=-337.18\text{kcal}$

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Chemistry 3 Marks Question