For a reaction at equilibrium-

1. $∆G=∆G^{\circ }=0$

2. $∆G=0 but not ∆G^{\circ }$

3. $∆G^{\circ }=0 but not ∆G$

4. $∆G=∆G^{\circ }\ne 0$

Using only the following data:

(I) $F{e}_2{O}_3\left(s\right)+3CO\left(g\right)\rightleftharpoons 2Fe\left(s\right)+3C{O}_2\left(g\right);$
$∆H^{\circ }=-26.8 kJ$

(II) $$\begin{gathered} Fe\left(s\right)+C{O}_2\left(g\right)\rightleftharpoons FeO\left(s\right)+CO\left(g\right); \\ ∆H^{\circ }=+16.5 kJ \end{gathered}$$

the $∆H^{\circ }$ value, in kilojoules, for the reaction

$F{e}_2{O}_3\left(s\right)+CO\left(g\right)\to 2FeO\left(s\right)+C{O}_2\left(g\right)$ is calculated to be:

1. -43.3

2. -10.3

3. +6.2

4. +10.3

82 litres of carbon dioxide are produced at a pressure of 1 atm by the action of acid on a metal carbonate. The work done at room temp by the gas (in calories) in pushing back the atmosphere is 

1. 1000

2. 820

3. 1640

4. 2200

Among the following, the reaction for which $∆H=∆E$ is-

1. $PC{l}_5\left(g\right)\to PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

2. $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$

3. $C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

4. $C_2{H}_4\left(g\right)+H_2\left(g\right)\to C_2{H}_6\left(g\right)$

For which change $∆H\ne ∆E$

1. $H_2+I_2\rightleftharpoons 2HI$

2. $HCl+NaOH\to NaCl+H_{2}O$

3. $C\left(s\right)+O_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)$

4. $N_2+3H_2\to 2N{H}_3$

Enthalpy change when 1.00 g water is frozen at $0^{\circ }C,$ is :

$\left(∆H_{fus}=1.435 kcal mo{l}^{-1}\right)$

1. 0.0797 kcal

2. -0.0797 kcal

3. 1.435 kcal

4. -1.435 kcal

One mole of hydrogen gas at ${25}^{\circ }C$ and 1 atm pressure is heated at constant pressure until its volume has doubled. Given that $C_v$ for hydrogen is 3.0 cal deg^-1 mol^-1, the $∆H$and $∆E$ for this process are-

1. $∆H$= 1490 cal and $∆E$ = 894 cal

2. $∆H$= $∆E$= 1490 cal

3. $∆H$= 894 cal and $∆E$= 1490 cal

4. $∆H$= $∆E$= 894 cal

The enthalpy change of the reaction

$A{l}_{2}C{l}_6\left(s\right)+6Na\left(s\right)\to 2Al\left(s\right)+6NaCl\left(s\right)$ is -257 kcal.

Given the heat of formation of NaCl(s) is 98.0 kcal, the heat of formation of $A{l}_{2}C{l}_6\left(s\right)$ is-

1. 158 kcal

2. -331 kcal

3. -166 kcal

4. 316 kcal

Some of the thermodynamic parameters are state variables while some are process variables. Some grouping of the parameters are given. Choose the correct one-

1. State variables : Temperature, No. of moles

Process variables : Internal energy, work done by the gas

2. State variables : Volume, Temperature

Process variables : Internal energy, work done by the gas

3. State variables : work done by the gas, heat rejected by the gas.

Process variables : Temperature, volume

4. State variables : Internal energy, volume

Process variables : Work done by the gas, heat absorbed by the gas

One mole of a real gas is subjected to heating at constant volume from$\left(P_1, V_1, T_1\right)$ state to $\left(P_2, V_2, T_2\right)$ state. Then it is subjected to irreversible adiabatic compression against constant external pressure of P₃ atm till syatem reaches the final state$\left(P_3, V_3, T_3\right)$. If the constant volume molar heat capacity of real gas is C_V. Find out correct expression for $∆H$ from state 1 to state 3-

1. $C_v\left(T_3-T_1\right)+\left(P_3{V}_1-P_1{V}_1\right)$

2. $C_v\left(T_2-T_1\right)+\left(P_3{V}_2-P_1{V}_1\right)$

3. $C_v\left(T_2-T_1\right)+\left(P_3{V}_1-P_1{V}_1\right)$

4. $C_p\left(T_2-T_1\right)+\left(P_3{V}_1-P_1{V}_1\right)$