Tag: chemistry

Questions Related to chemistry

Which of the following statements are correct?

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. absolute value of enthalpy cannot be determined.

  2. absolute value of internal energy cannot be determined.

  3. absolute value of entropy can be determined.

  4. internal energy, enthalpy, and entropy are intensive properties.

Show answer
Correct Option: A,B
Explanation:

Properties which depend on the amount of the substance (or substances)

present in the system are called extensive propterties. e.g. Mass,

volume, heat capacity, internal energy, entropy, Gibb's free energy (G),

surface area etc. These properties will change with change in the

amount of matter present in the system.
The absolute value of

internal energy cannot be determined because it is not possible to

determine the exact values for the constituent energies such as

translational, vibrational, rotational energies, etc. However, we can

determine the change in internal energy (U) of the system when it

undergoes a change from initial state (Ul) to final state (Uf).
Enthalpy

is not a matter. It doesn't have mass and it doesn't occupy space.

Therefore, we cannot measure its absolute value (like energy).
We can only measure the changes because the energy of the universe is constant.

Enthalpy of the system is given as

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $\,H + PV$

  2. $\,U + PV$

  3. $\,U - PV$

  4. $\,H - PV$

Show answer
Correct Option: B
Explanation:
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as :
$H=U+PV $ (By definition)
$\Delta H=\Delta U+\Delta (PV)$
$\Delta H=\Delta U+P\Delta V$ (at constant pressure) combining with first law.
$\Delta H=q _{p}$

Which of the following reactions have same heat of reaction at constant $P$ and constant volume as well?

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $2NO(g)\longrightarrow N _2(g)+O _2(g)$

  2. $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$

  3. $Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$

  4. $H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$

Show answer
Correct Option: A,C,D
Explanation:

The following reactions have the same heat of reaction at constant $P$ and constant volume as well.


$2NO(g)\longrightarrow N _2(g)+O _2(g)$
$Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
$H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products is the same.

$\because \Delta n _g = 0$

However, for the reaction $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$, heat of reaction at constant $P$ and constant volume are different.

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products are different.

For which reaction will $\Delta H = \Delta U$?


 Assume each reaction is carried out in an open container.

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $H _2(g) + Br _2(g)\longrightarrow 2HBr(g)$

  2. $C(s) + 2H _2O(g)\longrightarrow 2H _2(g) + CO _2(g)$

  3. $4CO(g) + 2O _2(g)\longrightarrow 4CO _2(g)$

  4. $2PCl _5(g)\longrightarrow 2PCl _3(g) + 2Cl _2(g)$

Show answer
Correct Option: A
Explanation:

$\Delta H = \Delta U + \Delta nRT$, where $\Delta n =$ Change in number of moles.


$H _2 + Br _2 \to 2HBr$


 $\Delta n = 0$.  

$\therefore$ $\Delta H = \Delta U$

Hence, option A is correct

In which of the following reactions, $\Delta H > \Delta U$?

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $H _2(g) + I _2(g)\rightarrow 2HI(g)$

  2. $PCl _5(g)\rightarrow PCl _3(g) + Cl _2(g)$

  3. $2H _2O _2(l)\rightarrow 2H _2O(l) + O _2(g)$

  4. $C(s) + O _2(g)\rightarrow CO _2(g)$

Show answer
Correct Option: B,C
Explanation:

As we know,
$\Delta H = \Delta U + \Delta nRT$, where $\Delta n = n _P - n _R$ (n = number of moles)
$H _2(g) + I _2(g)\rightarrow 2HI(g)$      $\Delta n = 0$
$PCl _5(g)\rightarrow PCl _3(g) + Cl _2(g)$                     $\Delta n = 1$
$2H _2O _2(l)\rightarrow 2H _2O(l) + O _2(g)$                 $\Delta n = 1$
$C(s) + O _2(g)\rightarrow CO _2(g)$                  $\Delta n = 0$

Enthalpy of the system is given as:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $U + PV$

  2. $H = PV$

  3. $U - PV$

  4. $ H - V$

Show answer
Correct Option: A
Explanation:

Enthalpy is a system is defined as,

$H=U+PV$
Hence, option A is correct

Match List I with List II and select the answer from the given codes. 


List I                                                                               List II
A. $C(s) + O _2(g)\longrightarrow CO _2(g)$                       1. $\Delta H = \Delta U + RT$
B. $N _2(g) + 3H _2(g)\longrightarrow 2NH _3(g)$                2. $\Delta H = \Delta U$
C. $NH _4HS(s)\longrightarrow NH _3(g) + H _2S(g)$       3. $\Delta H =\Delta U - 2RT$
D. $PCl _5(g)\longrightarrow PCl _3(g) + Cl _2(g)$              4. $\Delta H = \Delta U + 2RT$
E. $2SO _2(g) + O _2(g)\longrightarrow 2SO _3(g)$              5. $\Delta H = \Delta U - RT$

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $A-1, B-2, C-3, D-4, E-5$

  2. $A-5, B-2, C-3, D-4, E-1$

  3. $A-1, B-3, C-4, D-2, D-5$

  4. $A-2, B-3, C-4, D-1, E-5$

Show answer
Correct Option: D
Explanation:

As we know,


$\Delta H = \Delta U + \Delta nRT$
$C(s) + O _2(g)\longrightarrow CO _2(g)$       
$\Delta n =0$ so         $\Delta H = \Delta U$

$N _2(g) + 3H _2(g)\longrightarrow 2NH _3(g)$    
$\Delta n =0$ so  $\Delta H =\Delta U - 2RT$


$NH _4HS(s)\longrightarrow NH _3(g) + H _2S(g)$     
$\Delta n =+2$ so  $\Delta H =\Delta U + 2RT$

$PCl _5(g)\longrightarrow PCl _3(g) + Cl _2(g)$         
$\Delta n = 1$ so  $\Delta H =\Delta U + RT$

$2SO _2(g) + O _2(g)\longrightarrow 2SO _3(g)$
$\Delta n = -1 $ so  $\Delta H =\Delta U - RT$     

$H _2(g) + I _2(g)\longrightarrow 2HI(g)$ 


For this reaction, relate $\Delta H$ and $\Delta U$.

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $\Delta H$ =$\Delta U$

  2. $\Delta H$ > $\Delta U$

  3. $\Delta H$ < $\Delta U$

  4. None of these

Show answer
Correct Option: A
Explanation:

As we know,
$\Delta H$ =$\Delta U+ \Delta n _g RT$
here, 
$H _2(g) + I _2(g)\longrightarrow 2HI(g)$ 
$\Delta n _g = 0$
so,
$\Delta H$ =$\Delta U$

Heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction $N _2 + 3H _2 \longrightarrow  2NH _3$ differ $(\Delta H- \Delta U)$ by the amount:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $2RT$

  2. $-2RT$

  3. $3RT$

  4. $RT$

Show answer
Correct Option: B
Explanation:

The heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction 


$N _2 + 3H _2 \longrightarrow  2NH _3$

$\Delta H = \Delta U + (\Delta n _g)\times RT$

$\Delta H = \Delta U + (-2)\times RT$

This is because the change in the number of moles of gaseous products and the gaseous reactants in the above reaction is -2.

$\Delta H - \Delta U = (-2)\times RT$

Hence, option B is correct.

If $\Delta E$ is the heat of reaction for
${C} _{2}{H} _{5}OH\left(l\right) + 3{O} _{2}\left(g\right) \longrightarrow 2C{O} _{2}\left(g\right) + 3{H} _{2}O\left(l\right)$
at constant volume, the $\Delta H$ (heat of reaction at constant pressure), at constant temperature is:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $\Delta H = \Delta E + RT$

  2. $\Delta H = \Delta E - RT$

  3. $\Delta H = \Delta E - 2RT$

  4. $\Delta H = \Delta E + 2RT$

Show answer
Correct Option: B
Explanation:

We know that,
$\Delta H = \Delta E + \Delta nRT$
where,
$\Delta n =$ number of moles of gaseous products - number of moles of gaseous reactants
$= 2-3=-1$
So, $\Delta H = \Delta E - RT$