Exercise 2

The third law of thermodynamics states that in the limit $\mathrm{T}\to 0$,

For the reaction ${\mathrm{CaCO}}_3 \left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO} \left(\mathrm{s}\right)+{\mathrm{CO}}_2 \left(\mathrm{g}\right)$ at $927°\mathrm{C}$, $\mathrm{\Delta H}=176 \mathrm{kJ} {\mathrm{mol}}^{-1}$, then $∆\mathrm{E}$ is

Highest entropy is in

Given, $\mathrm{\Delta G}°=-{\mathrm{nFE}}_{\text{cell }}°$ and $\mathrm{\Delta G}°=-\mathrm{RT} \ln \mathrm{k}$. The value of $\mathrm{n}=\frac{\mathrm{F}}{\mathrm{RT}}$ will be given by the slope of which line in the figure

For spontaneous cooling of a hot metal rod (system) in the air,

Consider the following processes:

                                $\mathrm{\Delta H}(\mathrm{kJ}/\mathrm{mol})$

$\frac{1}{2} \mathrm{A}⟶\mathrm{B} +150$

$3\mathrm{B}⟶2\mathrm{C}+\mathrm{D} -125$

$\mathrm{E}+\mathrm{A}⟶2\mathrm{D} +350$

For $\mathrm{B}+\mathrm{D}⟶\mathrm{E}+2\mathrm{C}, \mathrm{\Delta H}$ will be:

$\mathrm{\Delta E}$ of combustion of isobutylene is $-\mathrm{X} \mathrm{kJ} {\mathrm{mol}}^{-1}$. The value of $\mathrm{\Delta H}$ is 

The value of $\mathrm{\Delta H}-\mathrm{\Delta E}$ in $\mathrm{J}$ for the following reaction at $27 °\mathrm{C}$ will be-

$2{\mathrm{NH}}_3\left(\mathrm{g}\right)⟶{\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)$