Embibe Experts Solutions for Chapter: Thermodynamics, Exercise 2: Level 2

A cylinder of cooking gas in a household contains $11.6 \mathrm{kg}$ of butane. The thermochemical reaction for the combustion of butane is $2{\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)+13{\mathrm{O}}_2\left(\mathrm{g}\right)⟶8{\mathrm{CO}}_2\left(\mathrm{g}\right)+10{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); \mathrm{\Delta H}=-2658 \mathrm{KJ}/\mathrm{mol}$. If the household needs $15000 \mathrm{KJ}$ of energy per day, the cooking gas cylinder will last for about:

For the reaction $\mathrm{A}\to \mathrm{B}, \mathrm{\Delta H}°=7.5 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $\mathrm{\Delta S}°=25 \mathrm{J} {\mathrm{mol}}^{-1}$, the value of $\mathrm{\Delta G}°$ and the temperature, at which the reaction reaches equilibrium are, respectively. 

Diborane is formed from the elements as shown in equation $\left(1\right)$.

Given that,  $2\mathrm{B}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{B}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)$    ...$\left(1\right)$

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) {\mathrm{\Delta H}}_1^0=44 \mathrm{kJ}$

$2\mathrm{B}+3/2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{B}}_2{\mathrm{O}}_3\left(\mathrm{s}\right) {\mathrm{\Delta H}}_2^0=-1273 \mathrm{kJ}$

${\mathrm{B}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{B}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) {\mathrm{\Delta H}}_3^0=-2035 \mathrm{kJ}$

${\mathrm{H}}_2\left(\mathrm{g}\right)+1/2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) {\mathrm{\Delta H}}_4^0=286 \mathrm{kJ}$

the ${\mathrm{\Delta H}}^0$ for the reaction $\left(1\right)$ is:

The $\mathrm{\Delta H}$ for vaporisation of a liquid is $20 \mathrm{kJ}/\mathrm{mol}$. Assuming ideal behaviour, the change in internal energy for the vaporization of $1$ mole of the liquid at $60 °\mathrm{C}$ and $1 \mathrm{bar}$ is close to:

Given that the bond energies of: $\mathrm{N}\equiv \mathrm{N}$ is $946 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{H}-\mathrm{H}$ is $435 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{N}-\mathrm{N}$ is $159 \mathrm{kJ} {\mathrm{mol}}^{-1}$, and $\mathrm{N}-\mathrm{H}$ is $389 \mathrm{kJ} {\mathrm{mol}}^{-1}$, the heat of formation of hydrazine in the gas phase in $\mathrm{kJ} {\mathrm{mol}}^{-1}$ is:

For the reaction ${\mathrm{N}}_2+3{\mathrm{X}}_2\to 2{\mathrm{NX}}_3$ where $\mathrm{X}=\mathrm{F}, \mathrm{Cl}$ (the average bond energies are $\mathrm{F}-\mathrm{F}=155 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{N}-\mathrm{F}=272 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{Cl}-\mathrm{Cl}=242 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{N}-\mathrm{Cl}=200 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $\mathrm{N}\equiv \mathrm{N}=941 \mathrm{kJ} {\mathrm{mol}}^{-1}$). The heats of the formation of ${\mathrm{NF}}_3$ and ${\mathrm{NCl}}_3$ in $\mathrm{kJ} {\mathrm{mol}}^{-1}$, respectively, are closest to:

In the reaction ${\mathrm{CS}}_2\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{SO}}_2\left(\mathrm{g}\right) \mathrm{\Delta H}=-265 \mathrm{kcal}$

The enthalpies of formation of ${\mathrm{CO}}_2$ and ${\mathrm{SO}}_2$ are both negative and are in the ratio $4:3$. The enthalpy of formation of ${\mathrm{CS}}_2$ is $+26 \mathrm{kcal}/\mathrm{mol}$. Calculate the enthalpy of formation of ${\mathrm{SO}}_2$.

In Haber's process of manufacturing of ammonia:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right); {\mathrm{H}}_{25°\mathrm{C}}^0=-92.2 \mathrm{kJ}$

If ${\mathrm{C}}_{\mathrm{p}}$ is independent of temperature, then reaction at $100°\mathrm{C}$ as compared to that of $25°\mathrm{C}$ will be: