Tag: study of enthalpy

Questions Related to study of enthalpy

Calorific value gives ___________.

chemistry occurrence of carbon compounds in nature calorific value study of enthalpy chemical thermodynamics

  1. fuel efficiency

  2. amount of heat

  3. amount of light

  4. none of the above

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Correct Option: A
Explanation:

Calorific value is a measure of heating capacity of fuel. It gives the fuel efficiency.

Which one of the following has the highest calorific value?

chemistry occurrence of carbon compounds in nature calorific value study of enthalpy chemical thermodynamics

  1. Kerosene

  2. Biogas

  3. LPG

  4. Petrol

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Correct Option: C
Explanation:

The calorific value of a fuel is the quantity of heat produced by its combustion - at constant pressure and under a conditions known as " normal " of temperature and pressure (i.e. to 0oC and under a pressure of 1,013 mbar). 
The combustion of a fuel product generates water vapor. Certain techniques are used to recover the quantity of heat contained in this water vapor by condensing it. 
The Higher Calorific Value (or Gross Calorific Value - GCV) supposes that the water of combustion is entirely condensed and that the heat contained in the water vapor is recovered. 
The Lower Calorific Value (or Net Calorific Value - NCV) supposes that the products of combustion contains the water vapor. The heat contained in the water vapor is not recovered. 
LPG has a typical specific calorific value of 46.1 MJ/kg.
Calorific value of Bio gas is about 19 to 25 MJ/cubic meter, depending on the methane percentage in the biogas mix.
Heat of combustion of kerosene for lower heating value is 43.1 MJ/kg, and for higher heating value is 46.2 MJ/kg.
Petrol, calorific value of which is 45.8 MJ/kg.
Thus, LPG has highest calorific value among above.

Which of the following compounds has the highest calorific value ?

chemistry occurrence of carbon compounds in nature calorific value study of enthalpy chemical thermodynamics

  1. $CH _4$

  2. $C _2H _6$

  3. $H _2$

  4. $C _4H _{10}$

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Correct Option: C
Explanation:

Calorific value is the amount of energy produced by the complete combustion of a molecule. Amongst the given options, $H _{2}$ has the highest calorific value.

Heat of reaction at constant pressure is called

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

  1. Enthalpy.

  2. Entropy

  3. Free energy

  4. None of these

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Correct Option: A
Explanation:

Enthalpy : Chemical reactions are generally carried out at constant pressure (atmospheric pressure) and heat of reaction at constant pressure is called Enthalpy (H) as:
$H = U + PV$ (By definition).

The relationship between enthalpy and internal energy change is

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

  1. $\,\Delta U = \Delta H + P\Delta V$

  2. $\,\Delta H = \Delta U + P\Delta V$

  3. $\,\Delta H = \Delta U - P\Delta V$

  4. $\,P\Delta V = \Delta U + \Delta H$

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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}$

An athelete is given 100 g of glucose $(C _6H _{12}O _6)$ of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, Determine the weight of water he would need to perspire. (The enthalpy of evaporation of water is 44 kJ/mole.)

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

  1. 319 gm

  2. 323 gm

  3. 342 gm

  4. 312

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Correct Option: A
Explanation:

 Net amount of energy given to athlete $= 1560 kJ$

$1560 × 50/100$ Energy lost in an event $= 780 kJ$
 Energy left out $= 1560 – 780 kJ = 780 kJ$
 Now, consider the evaporation of water $H _2O(l) \rightarrow H _2O(g)$; $\Delta H = 44 kJ mole^{–1}$
 Thus, for consumption of $44 kJ$ of energy the amount of water evaporated $=1mole= 18 g$ 
For consumption of $780 kJ$ of energy the amount of water to be evaporated $ \frac {18 \times  780}{44} = 319·09 g$

Enthalpy of the system is given as :

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

  1. $\displaystyle H+PV$

  2. $\displaystyle U+PV$

  3. $\displaystyle U-PV$

  4. $\displaystyle H-PV$

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Correct Option: B
Explanation:

Enthalpy: 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 :
$H = U + PV$ 

Under which of the following condition is the relation $\Delta H = \Delta U + P\Delta V$ valid for a closed system at

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

  1. constant pressure

  2. constant temperature

  3. constant temperature and pressure

  4. constant temperature, pressure and composition

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Correct Option: A
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 $

In the hydrogenation of ethylene,$\Delta n$ is equal to:

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

  1. 1

  2. -2

  3. -1

  4. 2

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Correct Option: C
Explanation:

$C _{2}H _{4}(g)+H _{2}(g)\rightarrow C _{2}H _{6}(g)$

$\Delta n=1-2$

$=$ - 1


Option C is correct.

For the reaction,

$C(s)+O {2}(g)\rightarrow CO _{2}(g);$

$\Delta n$ value is _________.

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

  1. Zero

  2. +1

  3. -1

  4. unpredictable

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Correct Option: A
Explanation:

$C(s)+O _{2}(g)\rightarrow CO _{2}(g)$


$\Delta n _{g}=1-1$

$=0$ 

$=$ difference of moles of gaseous substances

Option A is correct.