Tag: chemistry

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Multiple choice chemistry nuclei nuclear force the nuclear force nuclear force and binding energy

Consider an $\alpha$-particle just in contact with a $ _{\;  92}^{238}\textrm{U}$ nucleus. The Coulombic repulsion energy  (i.e, the height of the Coulombic barrier between $^{238}\textrm{U}$ and alpha particle) assuming that the distance between them is equal to the sum of their radii is 

  1. $16.35 \, MeV$
  2. $46.66 \, MeV$
  3. $22.24 \, MeV$
  4. $26.14 \, MeV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The expression for the radius of the nucleus is as shown below.
$r _{nucleus} =1.3\times 10^{-13}(A)^{1/3}$; where $A$ is mass number
Radius of  $ _{92}^{238}\textrm{U}=1.3\times10^{-13}\times (238)^{1/3}$
                           $= 8.06\times 10^{-13}cm$
Radius of $ _{2}^{4}\textrm{He}=1.3\times10^{-13}\times (4)^{1/3}$
                         $=2.06\times10^{-13}cm$
Total distance between uranium and helium nuclei is equal to the sum of their radii. 

It is $=(8.06 + 2.06)\times10^{-13}=10.12\times10^{-13}cm$ 

The Coulombic repulsion energy is: 
$\displaystyle \frac{Q _1Q _2}{r}$ $\displaystyle =\frac{92\times 4.8\times 10^{-10}\times 2\times 4.8\times 10^{-10}}{10.12\times 10^{-13}}erg$                (because $Q _1$ and  $Q _2$  in  esu and r in cm)     
                                      
            $=418.9\times 10^{-7}erg= 418.9\times 10^{-14}$J

            $=418.9\times 10^{-14}/1.602\times 10^{-19}\ eV$
              
            $\displaystyle =\frac{26.14\times 10^6}{10^6}\ MeV$

            $=26.14 \, MeV$

Hence, the coulombic repulsion energy is $26.14\ MeV$.

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

The relationship between enthalpy and internal energy change is

  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. $\,P\Delta V = \Delta U + \Delta H$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$
Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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.)

  1. 319 gm

  2. 323 gm

  3. 342 gm

  4. 312
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

Enthalpy of the system is given as :

  1. $\displaystyle H+PV$
  2. $\displaystyle U+PV$
  3. $\displaystyle U-PV$
  4. $\displaystyle H-PV$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 :
$H = U + PV$ 

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

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

  1. constant pressure

  2. constant temperature

  3. constant temperature and pressure

  4. constant temperature, pressure and composition
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $
Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

the $\triangle { H } _{ 1 }\quad $ and $\triangle { H } _{ 2 }\quad $ $(in kJ mol^-1)$ and the $\triangle {egH } $  $(in kJ mol^-1)$ of a few 3 elements are given below:
Which of the element is likel;y to be:

Elements $\triangle { H } _{ 1 }\quad $ $\triangle { H } _{ 2 }\quad $  $\triangle {egH } $
I 520 7300 -60
II 419 3051 -48
III 1681 3374 -328
IV 1008 1846 -295
V 2372 5251 +48
VI 738 1451 -40

  1. the least reactive element

  2. the most reactive mental

  3. the most reactive non- mental

  4. the least reactive non-mental
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Element V has a positive electron gain enthalpy (+48 kJ/mol), meaning it requires energy input to gain an electron. This makes it the least reactive as it resists electron gain. Element II (ΔH1=419, ΔH2=3051, ΔegH=-48) would be most reactive metal, and Element III (ΔH1=1681, ΔH2=3374, ΔegH=-328) would be most reactive non-metal.