Chemistry 5 Marks Questions

1. –74.8kJ mol^–1. 

Explanation:

According to the question,

1.  $\text{CH}_{4(\text{g})}+2\text{O}_{2(\text{g})}\xrightarrow{\ \ \ \ \ \ }\text{CO}_{2(\text{g})}+2\text{H}_2\text{O}_{(\text{g})}$

$\Delta\text{H}=-890.3\text{kJ} \ \text{mol}^{-1}$

2.  $\text{C}_{(\text{s})}+\text{O}_{{2(\text{g})}}\xrightarrow[]{\ \ \ \ \ \ \ }\text{CO}_{2(\text{g})}$

$\Delta\text{H}=-393.5\text{kJ} \ \text{mol}^{-1}$

3.  $2\text{H}_{2(\text{s})}+\text{O}_{{2(\text{g})}}\xrightarrow{\ \ \ \ \ \ \ }2\text{H}_2\text{O}_{(\text{g})}$

$\Delta\text{H}=-285.8\text{kJ} \ \text{mol}^{-1}$

Thus, the desired equation is the one that represents the formation of CH_4(g) i.e.,

$\text{C}_{(\text{s})}+2\text{H}_{{2(\text{g})}}\xrightarrow[]{\ \ \ \ \ \ \ }\text{CH}_{4(\text{g})}$

$\Delta_{\text{f}}\text{H}_{\text{CH}_4} \ =\Delta_{\text{c}}\text{H}_{\text{c}} + 2\Delta_{\text{c}}\text{H}_{\text{H}_2} - \Delta_{\text{c}}\text{H}_{\text{co}_2}$

$=[-393.5 +2(-258.8)-(-890.3)]\text{kJ} \ \text{mol}^{-1}$

$=-74.8\text{kJ} \ \text{mol}^{-1}$

$\therefore$ Enthalpy of formation of CH_4(g) = –74.8kJ mol^–1

1. $\log\text{K}=-\frac{\Delta\text{G}}{2.303\text{RT}}$

$\therefore\log\text{K}=-\frac{-33.2\times10^3\text{J mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$

$=5.82\ \text{or K}=6.6\times10^5$

2. For the given reaction, $\Delta\text{G}^\circ=\frac{1}{2}\times(-33.2)=-16.6\text{kJ mol}^{-1}$

$\log\text{K}=-\frac{-16.6\times10^3\text{J mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$

$=2.91\ \text{or K}=8.1\times10^2$

3. For the given reaction $\text{NH}_3(\text{g})\overrightarrow{\ \ \ \ \ }\ \frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})$

$\Delta\text{G}^\circ=-[-16.6\text{kJ mol}^{-1}]=16.6\text{kJ mol}^{-1}$

$\log\text{K}=-\frac{16.6\times10^3\text{JK}^{-1}\text{mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$

$=-2.91\text{ or K}=1.23\times10^{-3}$

1. $\Delta\text{S}$ = -ve because liquid is changing into solid.
2. $\Delta\text{S}$ = -ve because diamond has rigid structure.
3. $\Delta\text{S}$ = -ve because pressure is increasing, disorder will decrease.
4. $\Delta\text{S}$ = +ve because process is endothermic and spontaneous.
5. $\Delta\text{S}$ = +ve because gaseous products are being formed.

1. As heat capacity of water = heat capacity of calorimeter, the heat gained by water = heat lost by calorimeter.

= 125 × (296.5 - 286.4) × 4.184J = 5282J = 5.282kJ

2. $\Delta\text{H}=\text{q}_\text{p}$ and Cp is heat evolved or absorbed per mole for 1° fall or rise in temperature. Here, fall in temperature 320 - 293 = 27K.

Molar mass of CCl₂F₂ = 12 + 2 × 35.5 + 2 × 19

= 121g mol^-1

$\therefore$ Heat evolved from 1.25g if the sample on being cooled from 320K to 293K at constant pressure $=\frac{80.7}{121}\times1.25\times27\text{J}=22.51\text{J}$

Further, $\Delta\text{H}=\Delta\text{U}+\text{p}\Delta\text{V}=-22.51\text{J}$

$\Big[\therefore\text{p}\Delta\text{V}=1\text{atm}\times\frac{(248-274)}{1000}\text{L}=-0.026\text{L atm}\Big]$

$=-0.026\times101.325\text{J}=-263\text{J}$

$\therefore-22.51=\Delta\text{U}-2.63\text{J}$

$\Delta\text{U}=-22.51+2.63\text{J}=-19.88\text{J}$

1. Enthalpy of Neutralisation: Enthalpy of Neutralisation is defined as heat evolved when 1 mole of H from acid combines with 1 mole of base to form water.

It is because strong acids and bases are completely ionised in aqueous solution.

2. Law of conservation of energy is the basis of Hess's law.
3. $2\text{A}_2(\text{g})+5\text{B}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ 2\text{A}_2\text{B}_5(\text{g})$

$\Delta\text{n}=2-2-5=-5$

$\Delta\text{H}=-50.16\text{k}$

$\Delta\text{U}=?$

$\text{T}=273+27=300\text{K}$

$\Delta\text{H}=\Delta\text{U}+\Delta\text{nRT}$

$\Rightarrow\Delta\text{U}=\Delta\text{H}-\Delta\text{nRT}$

$\Rightarrow\Delta\text{U}=-50.16\text{kJ}-\frac{(-5)\times8.314\times300}{100}\text{kJ}$

$\Rightarrow\Delta\text{U}=-50.16\text{kJ}+\frac{12471}{1000}\text{kJ}$

$\Rightarrow\Delta\text{U}=-50.16\text{kJ}+12.471\text{kJ}$

​​​​​​​$\Rightarrow\Delta\text{U}=-37.689\text{kJ}$

1. If U₁ is initial internal energy, 'q' is heat supplied, w is work done on the system, then final internal energy $\text{U}_2=\text{U}_1+\text{q}+\text{w}$

$\text{U}_2-\text{U}_1=\text{q}+\text{w}$

$\Delta\text{U}=\text{q}+\text{w}$

2. $\text{H}_2(\text{g})+\text{Br}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{HBr(g)}$

$\Delta_\text{r}\text{H}^\circ=\text{B}_{\text{H}-\text{H}}+\text{B}_{\text{Br}-\text{Br}}-2\text{B}_{\text{H}-\text{Br}}$

$=436+192-2\times368$

$\Delta_\text{r}\text{H}^\circ=628-736$

$=-108\text{kJ mol}^{-1}$​​​​​​​

$\text{C}_\text{s}+2\text{H}_2\text{O}+^{\frac{1}{2}}\text{O}_{2(\text{g})}\xrightarrow{\ \ \ \ \ \ }\text{CH}_3\text{OH}_{(\text{I})}...(1)$ 

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

= (–393kJ mol^–1) + 2(–286kJ mol^–1) – (–726kJ mol^–1)

= (–393 – 572 + 726)kJ mol^–1

$\therefore\Delta_\text{f}\text{H}^\ominus[\text{CH}_3\text{OH}_{(\text{I})}]=-239\text{kJ} \ \text{mol}^{-1}$

1. 100g of glucose is equivalent to 1560kJ of energy

Energy utilised in the event $=\frac{1560\times50}{100}=780$

Energy left unutilised = 1560 - 780 = 780kJ

Enthalpy of vaporisation of water = 44kJ/ mol $=\frac{44}{18}\text{kJ/ g}$

$\therefore$ Water needed to perspire $=\frac{44}{18}\times780=1906.66\text{g}$

2. Adding Eq. (i) and (iii) and substract Eq. (ii),

$\text{C}_2\text{H}_4(\text{g})+3\text{O}_2(\text{g})+\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})-\text{C}_2\text{H}_6(\text{g})\\-\frac{7}{2}\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}+\text{H}_2\text{O(l)}\\-2\text{CO}_2(\text{g})-3\text{H}_2\text{O(l)}$

and $\Delta_\text{r}\text{H}^\circ=-1401\text{kJ mol}^{-1}-286.0\text{kJ mol}^{-1}\\-(-1550\text{kJ mol}^{-1})$

These equations give,

$\text{C}_2\text{H}_2(\text{g})+\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ }\ \text{C}_2\text{H}_6(\text{g});$ $\Delta_\text{r}\text{H}^\circ=-137\text{kJ mol}^{-1}$​​​​​​​

1. Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
2. Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
3. Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

$\Delta\text{H} = \Delta\text{U} + \Delta\text{n}_\text{g}\text{RT}$

Where,

$\Delta\text{U} =$ change in internal energy

$\Delta\text{n}_\text{g} =$ change in number of moles

For the given reaction,

$\Delta\text{n}_\text{g} = \sum\text{n}_\text{g} (\text{products}) – \sum\text{n}_\text{g} (\text{reactants})$

$= (2 – 1.5) \ \text{moles}$

$\Delta\text{n}_\text{g} =0.5 \ \text{moles}$

And,

$\Delta\text{U} = \ –742.7\text{kJ} \ \text{mol}^{–1}$

$\text{T} = 298\text{K}$

$\text{R} = 8.314 × 10^{–3}\text{kJ} \ \text{mol}^{–1} \ \text{K}^{–1}$

Substituting the values in the expression of $\Delta\text{H}:$

$\Delta\text{H} = (–742.7\text{kJ} \ \text{mol}^{–1}) + (0.5 \ \text{mol}) (298\text{K})\$8.314 × 10^{–3}\text{kJ} \ \text{mol}^{–1} \ \text{K}^{–1})$

$= –742.7 + 1.2$

$\Delta\text{H} = –741.5\text{kJ} \ \text{mol}^{–1}$

1. Standard enthalpy of formation is defined as enthalpy change when 1 mole of compound is formed from the constituting elements in their standard states.

1. Is not enthalpy of formation of CaCO₃ because it is not being formed from constituting elements.
2. Is not enthalpy of formation of HBr because 2 moles of HBr are being formed.

2. $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ \ }\text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$

$\Delta\text{H}=$ Bond energy of reactants - Bond energy of products

$=2\text{B}_{\text{O}=\text{O}}+4\text{B}_{\text{C}-\text{H}}-2\text{B}_{\text{C}=\text{O}}-4\text{B}_{\text{O}-\text{H}}$

$=2\times498+4\times414-2\times741-4\times464$

$=(996+1656-1482-1856)$

$=-686\text{kJ mol}^{-1}$

1. 'q’ is not a state function because it depends upon path.

'w' is not a state function because it depends upon path.

$\text{q}+\text{w}=\Delta\text{U}$ which is a state function because it is independent of path.

2. $\text{OF}_2(\text{g})+\text{H}_2\text{O(g)}\overrightarrow{\ \ \ \ \ \ \ }\ \text{O}_2(\text{g})+2\text{HF(g)}$

$\Delta\text{n}=1+2-2=1$

$\Delta_\text{r}\text{H}^\circ=\sum\Delta_\text{f}\text{H}^\circ(\text{products})-\sum\Delta_\text{f}\text{H}^\circ(\text{reactants})$

$=\Delta_\text{f}\text{H}^\circ(\text{O}_2)+2\Delta_\text{f}\text{H}^\circ(\text{HF})-\Delta_\text{f}\text{H}^\circ(\text{OF}_2)\\-\Delta_\text{f}\text{H}^\circ(\text{H}_2\text{O})$

$=0+2\times(-268.6)-23.0-(-241.8)$

$=537.2-23.0+241.8$

$=-318.4\text{kJ}$

$\Delta_\text{f}\text{H}^\circ=\Delta_\text{r}\text{U}^\circ+\Delta\text{nRT}$

$-318.4\text{kJ}=\Delta_\text{r}\text{U}^\circ+\frac{1\times8.314\times300}{1000}\text{kJ}$

$-318.4\text{kJ}=\Delta_\text{r}\text{U}^\circ+2.4942\text{kJ}$

$\Delta_\text{r}\text{U}^\circ=-320.8942\text{kJ}$

$\Delta\text{S}_\text{total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_{\text{s}\omega\text{r}}$

$\Delta\text{S}_{\text{s}\omega\text{r}}=\frac{\Delta\text{H}_{\text{s}\omega\text{r}}}{\text{T}}=\frac{\Delta\text{H}_\text{sys}}{\text{T}}$

$\Delta\text{S}_\text{total}=\Delta\text{S}_\text{sys}=\Big(-\frac{\Delta\text{H}_\text{sys}}{\text{T}}\Big)$

$\text{T}\Delta\text{S}_\text{total}=\text{T}\Delta\text{S}_\text{sys}-\Delta\text{H}_\text{sys}$

$\Delta\text{S}_\text{total}>0,\text{so}$

$\text{T}\Delta\text{S}_\text{sys}-\Delta\text{H}_\text{sys}>0$

$\Rightarrow-\big(\Delta\text{H}_\text{sys}-\text{T}\Delta\text{S}_\text{sys}\big)>0$

$-\Delta\text{G}>0$

$\Delta\text{H}_\text{sys}$ is the enthalpy change of a reaction, $\text{T}\Delta\text{S}_\text{sys}$ is the energy which is not available to do useful work, so, $\Delta\text{G}$ is the net energy available to do useful work and is thus a measure of the 'free energy'. For this reason, it is also known as the free energy of the reaction.

$\Delta\text{G}$ gives a criteria for spontaneity at constant pressure and temperature, 

1. If $\Delta\text{G}$ is negative (<0), the process is spontaneous.

2. If $\Delta\text{G}$ is positive (>0), the process is non - spontaneous.

• Unit of $\Delta\text{G}$ is Joule.

• The reaction will be spontaneous at high temperature.

$\Delta\text{n}_\text{g}=2-(3)$

$\Delta\text{H}=\Delta\text{U}+\text{P}\Delta\text{V}$

$\Delta\text{H}=\Delta\text{U}+\text{P}(0)$

$\Delta\text{H}=\Delta\text{U}$

$\text{Pv}=\text{nRT}$

$\text{pv}_\text{A}=\text{n}_\text{A}\text{RT}$

$\text{pv}_\text{B}=\text{n}_\text{B}\text{RT}$

$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ(\text{Na}_2\text{CO}_3)+\Delta_\text{f}\text{H}^\circ(\text{CO}_2)\\+\Delta_\text{r}\text{H}^\circ(\text{H}_2\text{O})-2\Delta_\text{f}\text{H}^\circ(\text{NaHCO}_3)$

$\Delta_\text{r}\text{S}^\circ=\Delta_\text{r}\text{S}^\circ_\text{m}(\text{Na}_2\text{CO}_3)+\text{S}^\circ_\text{m}(\text{CO}_2)\\+\text{S}^\circ_\text{m}(\text{H}_2\text{O})-2\text{S}^\circ_\text{m}(\text{NaHCO}_3)$

$\Delta_\text{r}\text{S}^\circ=\frac{\Delta_\text{r}\text{H}^\circ}{\text{T}}$

$\therefore\text{T}=\frac{\Delta_\text{r}\text{H}^\circ}{\Delta_\text{r}\text{S}^\circ}=\frac{129.17}{334.37\times10^{-3}}=386.3\text{K}$

$\therefore$ Reaction will be spontaneous above 386.3K.

1.  $\text{CCl}_{4(\text{l})}\xrightarrow{ \ \ \ \ \ }\text{CCl}_{4(\text{g})}\Delta_\text{vap}\text{H}^\ominus=30.5\text{kJ} \ \text{mol}^{-1}$

2.  $\text{C}_{(\text{s})}\xrightarrow[]{\ \ \ \ \ \ \ }\text{C}_{(\text{g})} \ \Delta_{\text{a}}\text{H}^\ominus=715.0\text{kJ} \ \text{mol}^{-1}$

3.  $\text{Cl}_{2(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ }2\text{Cl}_{(\text{g})} \ \Delta_{\text{a}}\text{H}^\ominus=242\text{kJ} \ \text{mol}^{-1}$

4.  $\text{C}_{(\text{g})}+4\text{Cl}_{(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ }\text{CCl}_{4(\text{g})} \ \Delta_{\text{f}}\text{H}=-135.5\text{kJ} \ \text{mol}^{-1}$

Enthalpy change for the given process $\text{CCI}_{4(\text{g})}\xrightarrow{ \ \ \ \ \ }\text{C}_{(\text{g})}+4\text{Cl}_{(\text{g})}$, can be calculated

using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

$\Delta\text{H}=\Delta_\text{a}\text{H}^\ominus(\text{C})+2\Delta_\text{a}\text{H}^\ominus(\text{Cl}_2)-\Delta_\text{vap}\text{H}^\ominus-\Delta_\text{f}\text{H} $

$= (715.0\text{kJ} \ \text{mol}^{–1}) + 2(242\text{kJ} \ \text{mol}^{–1}) – (30.5\text{kJ} \ \text{mol}^{–1}) – (–135.5\text{kJ} \ \text{mol}^{–1})$

$\therefore\Delta\text{H}=1304\text{kJ} \ \text{mol}^{-1}$

Bond enthalpy of C–Cl bond in CCl_4(g)

$=\frac{1304}{4}\text{kJ} \ \text{mol}^{-1}$

$=326\text{kJ} \ \text{mol}^{-1}$

Dividing in half a tank of air at a given T,P and mass yields two smalller tanks containing air, but the mass of the air in each new, smaller tank is half as it was initially in the one big tank.

Internal energy: Extensive

Dividing in half a tank of water at a given T,P with a given total internal energy yields two smalller tank containing water at the same T and P, but the internal energy of the water in each new, smaller tank is half as it was initially in the one big tank.

Pressure: Intensive

The ratio of two extensive properties is also intensive

$\text{Pressure}=\frac{\text{Force}}{\text{Area}}=\frac{\text{Extensive}}{\text{Extensive}}=\text{Intensive}$

Heat capacity: Extensive

A sample containing twice the amount of substance as another sample requires the transfer of twice the amount of heat to achieve the same change in temperature.

Density: Intensive

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{Extensive}}{\text{Extensive}}=\text{Intensive}$

Mole fraction: Intensive

$\text{Mole}\ \text{fraction}=\frac{\text{Moles}}{\text{Total}\ \text{moles}}=\frac{\text{Extensive}}{\text{Extensive}}=\text{Intensive}$

Specific heat: Intensive

$\text{Specific}\ \text{heat}=\frac{\text{Heat}\ \text{capacity}}{\text{Mass}}=\frac{\text{Extensive}}{\text{Extensive}}=\text{Intensive}$

Temperature: Intensive

Dividing in half a tank of water at a given temperature yields two smaller tanks containing water at the same temperature as that in the big tank.

Molarity: Intensive

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}=\frac{\text{Extensive}}{\text{Extensive}}=\text{Intensive}$

1. $\Delta\text{S}=+\text{ve}$
2. $\Delta\text{S}=-\text{ve}$

1. Gibb's energy: It is defined as energy which can be converted into useful work.

$\text{G}=\text{H}-\text{TS}$

$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$

$\Delta\text{G}=-\text{ve}$ for spontaneous process

• Standard enthalpy of combustion: Standard enthalpy of combustion is the amount of heat evolved when one mole of a substance under standard condition is completely burnt to form product also under standard conditions.

• Standard enthalpy of formation: Standard enthalpy of formation is the enthalpy change accompanying the formation of one mole of a substance from its constituent element in their standard state, e.g. standard enthalpy of formation of CO₂ may be represented as C(s) + O₂(g) → CO₂(g)

1. Molecular formula of butane = C₄H₁₀

Molecular mass of butane = 4 × 12 + 10 × 1 = 58

Heat of combustion of butane = 2658kJ mol^-1

1 mole or 58g of butane on complete combustion gives heat = 2658kJ

$\therefore$ 14 × 10³g of butane on complete combustion will gives heat $=\frac{2658\times14\times10^3}{58}=641586$

The family needs 20000kJ of heat per day.

$\because$ 20000kJ of heat is used for cooking by a family in 1 day.

$\therefore$ 641586 kJ of heat will be used for cooking by a family in $=\frac{641586}{20000}=32\text{days}$

The cylinder will last for 32 days.

2. 25 percent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by 75% combustion of butane $=\frac{641586\times75}{100}=481189.5\text{kJ}$

$\therefore$ The number of days the cylinder will last.

$\text{w}=-2.303\text{nRT}\log\frac{\text{V}_2}{\text{V}_1}$