Hess' Law

Q.

Ethyl chloride $\left(C_2{H}_{5}Cl\right)$, is prepared by reaction of ethylene with hydrogen chloride :

$C_2{H}_4\left(g\right)+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right)\Delta H=-72.3kJ$. What is the value of $△E$ (in KJ), if 70g of ethylene and 73g of HCL are allowed to react at 300K.

1. −69.8

2. −174.5

3. −180.75

4. −139.6

Q.

Given the bond energies $N\equiv N$, $H-H$ and $N-H$ bonds are $945,436$ and $391kJmol{e}^{-1}$ respectively, the enthalpy of the following reaction $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is

1. 102 kJ

2. 105 kJ

3. 90 kJ

4. -93 kJ

Q. 10.C+H2O=CO+H2 . H20 ACTS AS AN OXIDISING AGENT WHY?

Q. The standard heat of formation of $F{e}_2{O}_3\left(s\right)$ is -824 kJ $mo{l}^{-1}$. The heat change for reaction $4Fe\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$ is:

1. -1648.4 kJ
2. - 824.2 kJ
3. 648.4 kJ
4. 1648.4 kJ

Q.

If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ $mo{l}^{-1}$, the value of enthalpy change for the reaction

1. -250 kJ

2. +125 kJ

3. -125 kJ

4. +250 kJ

Q. Calculate the lattice enthalpy of $KCl$ from the following data at standard states-
Enthalpy of sublimation of $K=89kJmo{l}^{-1}$
Enthalpy of dissociation of $C{l}_2=244kJmo{l}^{-1}$
Ionisation energy of $K=425kJmo{l}^{-1}$
Electron gain enthalpy of $Cl=-355kJmo{l}^{-1}$
Enthalpy of formation of $KCl=-438kJmo{l}^{-1}$

1. $719kJmo{l}^{-1}$
2. $-719kJmo{l}^{-1}$
3. $-832kJmo{l}^{-1}$
4. $832kJmo{l}^{-1}$

Q.

Why does Hesss law hold true?

Q. Hess law is applicable for the determination of heat of

1. Reaction
2. Formation
3. Transition
4. All of these

Q. The enthalpy change for a reaction $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

1. -92.2 kJ/mol
2. -46.1 kJ/mol
3. 46.1 kJ/mol
4. -92.2 kJ/mol

Q.

Given the bond energies $N\equiv N$, $H-H$ and $N-H$ bonds are $945,436$ and $391kJmol{e}^{-1}$ respectively, the enthalpy of the following reaction $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is

1. -93 kJ

2. 102 kJ

3. 90 kJ

4. 105 kJ

Q. The enthalpy of solution of $BaC{l}_2\left(s\right)$ and $BaC{l}_2.2H_{2}O\left(s\right)$ are – 20.6 and 8.8 kJ $mo{l}^{-1}$ respectively, the enthalpy change for the hydration of $BaC{l}_2\left(s\right)$ is:

1. 29.4 kJ
2. – 29.4 kJ
3. – 11.8 kJ
4. + 11.8 kJ

Q. A: HNO, acts as reducing agent onlyR: HNO, oxidises to HNO only but not reduce byany reducing agent.

Q.

The standard heat of formation values of $S{F}_6\left(g\right)$ , $S\left(g\right)$ and $F\left(g\right)$ are: $-1100,275$ and $80KJmo{l}^{-1}$ respectively. Then the average S - F bond energy in $S{F}_6$ is:

1. $301KJmo{l}^{-1}$

2. $-183KJmo{l}^{-1}$

3. $309KJmo{l}^{-1}$

4. $280KJmo{l}^{-1}$

Q. Hess' law is used

1. When a directly measured enthalpy change of reaction is not available
2. To calculate an enthalpy change value through multiple steps.
3. Because enthalpy is a state function.
4. All of these options are correct

Q.

The standard heat of formation values of $S{F}_6\left(g\right)$ , $S\left(g\right)$ and $F\left(g\right)$ are: $-1100,275$ and $80KJmo{l}^{-1}$ respectively. Then the average S - F bond energy in $S{F}_6$ is:

1. $301KJmo{l}^{-1}$

2. $-183KJmo{l}^{-1}$

3. $309KJmo{l}^{-1}$

4. $280KJmo{l}^{-1}$

Q. Calculate the lattice enthalpy of $KCl$ from the following data at standard states-
Enthalpy of sublimation of $K=89kJmo{l}^{-1}$
Enthalpy of dissociation of $C{l}_2=244kJmo{l}^{-1}$
Ionisation energy of $K=425kJmo{l}^{-1}$
Electron gain enthalpy of $Cl=-355kJmo{l}^{-1}$
Enthalpy of formation of $KCl=-438kJmo{l}^{-1}$

1. $719kJmo{l}^{-1}$
2. $-719kJmo{l}^{-1}$
3. $-832kJmo{l}^{-1}$
4. $832kJmo{l}^{-1}$

Q.

Ethyl chloride $\left(C_2{H}_{5}Cl\right)$, is prepared by reaction of ethylene with hydrogen chloride :

$C_2{H}_4\left(g\right)+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right)\Delta H=-72.3kJ$. What is the value of $\Delta E$ (in KJ), if 70g of ethylene and 73g of $HCl$ are allowed to react at 300K?

1. −69.8

2. −180.75

3. −174.5

4. −139.6

Q. If X is the lattice enthalpy of $CaC{l}_2$, find X. Given that the enthalpy of
(i) Sublimation of $Ca\left(s\right)$ is $121kJmo{l}^{-1}$
(ii) Dissociation of $C{l}_2\left(g\right)$ to $2Cl\left(g\right)$ is $242.8kJmo{l}^{-1}$
(iii) Ionization of $Ca\left(g\right)$ to $C{a}^{2+}\left(g\right)$ is $2422kJmo{l}^{-1}$
(iv) Electron gain enthalpy for $Cl\left(g\right)$ to $C{l}^{-1}\left(g\right)$ is $-355kJmo{l}^{-1}$
(v) ${△}_{f}H$ overall is $-795kJmo{l}^{-1}$

1. 2870.8 kJ/mol
2. -2372.6 kJ/mol
3. -2870.8 kJ/mol
4. 2372.6 kJ/mol

Q.

KP for the reaction $N_2+3H_2\leftrightharpoons 2N{H}_3$ at ${400}^{\circ }C$ is $1.64\times {10}^{-4}$. Calculate $K_c$.

1. 0.3 mole2 litre-2

2. 0.4 mole2litre-2

3. 0.5 mole2 litre-2

4. 0.6 mole2 litre-2

Q. The law of equivalents works only for a reaction going to completion so, what happens to the reaction that is not going to completion and how are they equilibriated?

Q. Determine the heat of transformation of C(diamond)→ C(graphite) from the following data.

$C_{\left(diamond\right)}+O_2\left(g\right)\to C{O}_2\left(g\right)△H=-94.5Kcal$
$C_{\left(graphite\right)}+O_2\left(g\right)\to C{O}_2\left(g\right)△H=-94.0Kcal$

1. -188.5 K cal

2. 188.5 K cal

3. -0.5 K cal

4. 0.5 K cal

Q. Calculate the reasonance energy of $N_{2}O$ from the following data.of $△H_f^o$ $N_{2}O=82kJmo{l}^{-1}$

Bond energy of N $\equiv$ N, N = N, O = O and N = O bonds are 946, 418, 498 & 607 $kJmo{l}^{-1}$ respectively.

1. 80 kJ

2. 85 kJ

3. 88 kJ

4. 90 KJ

Q. In $C{H}_4$ molecule, all $C-H$ bonds of methane have same energy.

1. True
2. False

Q. Hess' law is applicable for determination of enthalpy of

1. Reaction
2. Formation
3. Transition
4. All of the above

Q.

In a neutral or faintly alkaline solution, $8$ moles of permanganate anion quantitatively oxidized thiosulphate anions to produce $\mathrm{X}$ moles of a sulfur-containing product. The magnitude of $\mathrm{X}$ is

Q.

The standard heat of formation values of $S{F}_6\left(g\right)$ , /S(g) and $F\left(g\right)$ are: $-1100,275$ and $80KJmo{l}^{-1}$ respectively. Then the average S - F bond energy in $S{F}_6$ is:

1. 301 KJ mol^-1

2. 320 KJ mol^-1

3. 309 KJ mol^-1

4. 280 KJ mol^-1

Q. Calculate the $△$H of the reaction

$H-\underset{\stackrel{|}{Cl}}{\stackrel{\underset{|}{H}}{C}}-Cl\left(g\right)⟶C\left(s\right)+2H\left(g\right)+2Cl\left(g\right)$

Bond energy for C-H bond & C-Cl bond are 415KJ and 326KJ respectively.

1. -1482 kJ

2. 1482 kJ

3. 1483 kJ

4. -1483 kJ

Q.

Write the born-haber cycle for the formation of cacl2

Q.

The molar entropies of $HI\left(g\right),H\left(g\right)$ and $I\left(g\right)$ at $298K$ are $206.5,114.6$, and $180.7Jmol{e}^{-1}{K}^{-1}$ respectively. Using the $\Delta G^0$ given below, Calculate the bond energy of HI.

$H\left(g\right)\to H\left(g\right)+I\left(g\right)$; $\Delta G^0=271.8kJ$

1. 282.4 kJ/mole

2. 298.3 kJ/mole

3. 32.17kJ/mole

4. 100 kJ/mole

Q. Determine the heat of transformation of C(diamond)→ C(graphite) from the following data.

$C_{\left(diamond\right)}+O_2\left(g\right)\to C{O}_2\left(g\right)△H=-94.5Kcal$

$C_{\left(graphite\right)}+O_2\left(g\right)\to C{O}_2\left(g\right)△H=-94.0Kcal$

1. -0.5 K cal

2. 0.5 K cal

3. 188.5 K cal

4. -188.5 K cal