• Change in enthalpy in different processes / reactions
• Enthalpy of reaction
• Enthalpy of formation
• Enthalpy of phase transition
• Enthalpy of other processes / reactions

• Entropy
• Gibbs free energy

Some processes are spontaneous and some are non spontaneous

A spontaneous process means that the process has a tendency/potential to occur without any external assistance

Reverse spontaneous → non spontaneous

Decrease in energy of the system is not a criteria for a process to happen spontaneously

• Extensive
• state fuction
• $\Delta S$ - does not depend on path

• q>0$\quad$ $\Delta S_{system}>0$
• same amont of q at lower temp $\Delta S$> at higher temp
• $\Delta S∝\frac{1}{T}$

$\Delta S_{surr} = \frac{-q}{T}$ $\quad$ $\Delta S = \frac{q_{rev}}{T}$

adiabatic wall

q = 0

$\Delta S_{surr}$ = 0

adiabatic process

$(P_1,T_1,V_1)(I) \rightarrow (P_2,T_2,V_2)(II)$

from given information state I and state II

$q_{rev}$, $\Delta S = \frac{q_{rev}}{T}$

$\Delta S_{Total}>0$ $\Delta S + \Delta S{surr}>0$ $\Delta S -\frac{q}{T_{surr}}$>0

Sysytem and surronding are in thermal equilibrium $\Delta T_{surr} = {T}$ $\Delta S - \frac{q}{T}>0$

process is constant

$q_p=\Delta H$

$\Delta S - \frac{\Delta H}{T}>0$

$\Delta H - T\Delta S<0$

Temp is constant

$\Delta H - \Delta (TS)<0$

$\Delta (H-TS)<0$

• constant temperature and pressure-

• For a chemical reaction -

$\Delta rG^0 = -13.6 KJ mol^{-1}$ at 298K

log K = $-\frac{\Delta rG^0}{2.303RT}$

Choose the correct answer

A. Thermodynamic state function is a quantity

(i) Used to determine heat changes

(ii) Whose value is independent of path

(iii) Used to determine pressure volume work

(iv) Whose value depends on temperature only

Answer. is (ii)whose value is independent of path

For the process to occur under adiabatic conditions, the correct condition is:

$\Delta T = 0$ $\hspace{2mm}$(ii) $\Delta p$ = 0$\hspace{2mm}$(iii) q= 0 $\hspace{2mm}$(iv) w= 0

Answer. is (iii) q= 0

Standard enthalpies of formation of all elements in their reference states

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

$\Delta U$ of combustion of methane is $-$ X KJ$mol^{-1}$. The value of $\Delta H$ is

(i) = $\Delta U$

(ii) $>\Delta U$

(iii) $<\Delta U$

(iv) = 0

Solution

$CH_4(s) + 2O_2(g) \rightarrow CO_2(s) + 2H_2O(l)$

$\Delta n_g$ = -2

$\Delta H = \Delta U + \Delta n_g$ R T

$\Delta u$- 2 RT

Answer is (iii)$<\Delta U$

A reaction. A + B $\rightarrow$ C + D + q is found to have a positive entropy change.The reaction will be

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(iv) possible at any temperature

Solution

$\Delta H <0$

$\Delta S >0$

$\Delta G = \Delta H - T\Delta S$

In a process 701 J of heat is absorbed by a system and 394 J of work is done by the system. what is the change in internal energy for the process?

Solution

q = 701 J

w = - 394J

$\Delta u$ = q + w

calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 C to ice at 10.0C.$\Delta _{fus}H =6.03 kJ mol^{-1}$ at 0 C

$\Delta C_p{H_2O(l)} = 75.3J mol^{-1} K^{-1}$

$\Delta C_p{H_2O(s)} = 36.8 J mol^{-1} K^{-1}$

Solution

$10^\circ C$ water$\rightarrow0^\circ$ C water = $C_p, H_2O(l) \times \Delta T$

water$\rightarrow$ ice at $0^\circ C = -6.03 kJ mol^{-1}$

$0^\circ$ C ice $\rightarrow 10^\circ C$ ice = $C_p, H_2O(s)\times \Delta T$

Enthalpies of formation of $CO(g)$ , $CO_2(g)$,$N_2O$ and $N_2O_4(g)$ are 110, 393. 81 and 9.7 kJ $mol^{-1} respectively$. Find the value of$\Delta _rH$ for the reaction

$N_2O_4 (g)+ 3CO(g)\rightarrow N_2O(g)+ 3CO_2(g)$

Solution

• $\Delta _rH =∑ a_i \Delta _fH^0(product) -∑ b_i\Delta _fH^0(reactant)$

Given

$N₂(g) + 3H_2(g) → 2NH_3 (g) : \Delta _rH = -92.4 kJ mol^{-1}$

what is the standard enthalpy of formation of $NH_3$ gas?

T = 298K

$\Delta fH^0 = \frac{1}{2}(-92.4)KJ mol^{-1}$