Enthalpy Change and Types of Thermochemical Reactions

From the data of following bond energies:

 $\begin{array}{l}H-Hbondenergy:431.37kJmo{l}^{-1}\\ C=Cbondenergy:606.10kJmo{l}^{-1}\\ C-Cbondenergy:336.49kJmo{l}^{-1}\\ C-Hbondenergy:410.50kJmo{l}^{-1}\end{array}$

Calculate the enthalpy of the following reaction in $kJmo{l}^{-1}$.

Bond dissociation enthalpy of  $H_2,C{l}_{\mathrm{2 }}\mathrm{and}HCl\mathrm{ are }434,242\mathrm{ and} \mathrm{431 }\mathrm{k}\mathrm{J}\mathrm{ m}\mathrm{o}{\mathrm{l}}^{-1}$ respectively. The enthalpy of formation of HCl is:  

Standard heat of formation of ${\mathrm{SF}}_6\left(\mathrm{g}\right)$, $\mathrm{S}\left(\mathrm{g}\right)$ and $\mathrm{F}\left(\mathrm{g}\right)$ are: $-1100,$ $275$ and $80 \mathrm{kJ} {\mathrm{mol}}^{-1}$, respectively. Then, the average value of $\mathrm{S}-\mathrm{F}$ bond energy in ${\mathrm{SF}}_6$ is

If ${\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)\left(\mathrm{g}\right)=-85 \mathrm{kJ} {\mathrm{mol}}^{-1}$, ${\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{C}}_3{\mathrm{H}}_8\right)\left(\mathrm{g}\right)=-104 \mathrm{kJ} {\mathrm{mol}}^{-1}, \mathrm{\Delta H}°$  for $\mathrm{C}\left(\mathrm{s}\right)\to \mathrm{C}\left(\mathrm{g}\right)$ is $718 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and heat of formation of $\mathrm{H}-$atom is  $218 \mathrm{kJ} {\mathrm{mol}}^{-1}$, then:

Calculate $\mathrm{\Delta H}$ (in joules) for ${\mathrm{C}}_{\left(\mathrm{graphite}\right)}\to {\mathrm{C}}_{\left(\mathrm{diamond}\right)}$ from the following data,

${\mathrm{C}}_{(\mathrm{graphite}) }+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}; \mathrm{\Delta H}=-393.5 \mathrm{kJ}$

${\mathrm{C}}_{(\mathrm{diamond}) }+{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}; \mathrm{\Delta H}=-395.4 \mathrm{kJ}$

Enthalpy of vaporisation can be defined as the enthalpy change that accompanies the vaporisation of ten moles of liquid at any temperature and pressure.

Nitroglycerine $\left(\mathrm{MW}=227.1\right)$ detonates according to the following equations,

$2{\mathrm{C}}_3{\mathrm{H}}_5{\left({\mathrm{NO}}_3\right)}_3\left(l\right)⟶3 {\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+6{\mathrm{CO}}_2\left(\mathrm{g}\right)+5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The standard molar enthalpies of formation, ${\mathrm{\Delta H}}_{\mathrm{f}}°$ for all the compounds are given below.

${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{C}}_3{\mathrm{H}}_5{\left({\mathrm{NO}}_3\right)}_3\right]=-364 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{CO}}_2\left(\mathrm{g}\right)\right]=-393.5 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right]=-241.8 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{N}}_2\left(\mathrm{g}\right)\right]=0 \mathrm{kJ}/\mathrm{mol}$
${\mathrm{\Delta H}}_{\mathrm{f}}°\left[{\mathrm{O}}_2\left(\mathrm{g}\right)\right]=0 \mathrm{kJ}/\mathrm{mol}$

The enthalpy change when $10 \mathrm{g}$ of nitroglycerine is detonated is

You want to split carbon dioxide (${\mathrm{CO}}_2$) into carbon and oxygen. If energy from coal were applied to drive the decomposition reaction, less ${\mathrm{CO}}_2$ would be released than consumed, because no process is perfectly efficient.

From the following data
${\mathrm{CH}}_3{\mathrm{OH}}_{\left(l\right)}+\frac{3}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}; {\mathrm{\Delta }}_{\mathrm{r}}\mathrm{H}°=-726 \mathrm{kJ} {\mathrm{mol}}^{-1}$
${\mathrm{H}}_{2\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)}; {\mathrm{\Delta }}_{\mathrm{r}}\mathrm{H}°=-286 \mathrm{kJ} {\mathrm{mol}}^{-1}$
${\mathrm{C}}_{\left(\mathrm{graphite}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}; {\mathrm{\Delta }}_{\mathrm{r}}\mathrm{H}°=-393 \mathrm{kJ} {\mathrm{mol}}^{-1}$
The standard enthalpy of formation of ${\mathrm{CH}}_3{\mathrm{OH}}_{\left(l\right)}$ in $\mathrm{kJ} {\mathrm{mol}}^{-1}$ is

Write a thermochemical equation to depict standard enthalpy of sublimation ?

Standard enthalpy of _____ is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure.

A container holds $2$ moles of gas $\mathrm{A}$ and $3$ moles of gas $\mathrm{B}$ in it at a constant pressure of $2 \mathrm{bar}$. The gases are separated by a wall and the whole container is held in an ice bath at a constant temperature of $0°\mathrm{C}$. The wall is removed and the gases mix and react to form $\mathrm{C}$ by the reaction given below. After the reaction comes to equilibrium how many grams of the ice have melted ? Consider that all heat produced in reaction has been utilized in melting of ice.

$\text{A}\left(\text{g}\right)+\text{B}\left(\text{g}\right)\rightleftharpoons \text{C}\left(\text{g}\right)$

All data above is at $298 \mathrm{K}$. $\Delta {}_{\text{FUS}}{\text{H}}^{\circ }\left(\text{water, 0}\mathrm{℃}\right)={\text{6 kJ mol}}^{-1}$.

Read the passage and answer the following questions as single choice correct .$\left(\text{Given 273}\times \text{8.314}\times \text{2.303}={\text{5227 J m}}^{-1}\right)$

Mass of ice melted in the reaction may be

Given that $\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2 , ∆\mathrm{H}° = - \mathrm{x} \mathrm{kJ}$

$2\mathrm{CO} + {\mathrm{O}}_2\to 2{\mathrm{CO}}_2 , ∆\mathrm{H}°= - \mathrm{y} \mathrm{kJ}$

The enthalpy of formation of carbon monoxide will be

Which of the following reaction defines $∆{{\mathrm{H}}_{\mathrm{f}}}^{°}$?

In the reaction : $\mathrm{S}+3/2\mathrm{ }{\mathrm{O}}_2\to \mathrm{S}{\mathrm{O}}_3+2\mathrm{x}$ kcal and $\mathrm{S}{\mathrm{O}}_2+1/2{\mathrm{O}}_2\to \mathrm{S}{\mathrm{O}}_3+\mathrm{y}$ kcal,
heat of formation of $\mathrm{S}{\mathrm{O}}_2$ is