Tag: chemistry

Questions Related to chemistry

Law of multiple proportions is illustrated by which one of the  following pairs ?

chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

  1. $H _{2}S$ and $SO _{2}$

  2. $NH _{3}$ and $NO _{2}$

  3. $N _{2}O$ and NO

  4. $N _{2}S$ and $Na _{2}O$

Show answer
Correct Option: C
Explanation:
Law of multiple proportions say that if 2 elements form more than one compounds between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.
In $N _{2}O$ and $NO$, the ratio of weight of $N$ reacting with a constant weight of $O$ is $2:1$

State True or False.
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.

chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

When two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio. This is the law of multiple proportions.
Consider two elements $C$ and $O$. They combine to form $CO$ and $CO _2$.
The masses of oxygen that combine with $12$ g of carbon to form $CO$ and $CO _2$ are $16$ g and $32$ g respectively. They are in whole number ratio $1:2$.

$P _xO _y$ what will be correct value of $x$ and $y$ if $P$ and $H$ combine in the mass ratio of $3.1 : 0.3$ and in water $H$ and $O$ combine in the mass ratio $0.2 : 1.6$?

chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

  1. $1$ and $1$

  2. $1$ and $2$

  3. $2$ and $3$

  4. $2$ and $5$

Show answer
Correct Option: C
Explanation:

 

$P:H = 3.1 : 0.3 = (3.1 \times : 0.30) \times 2 = 6.2 : 0.6$

$H:O = 0.2 : 1.6 = (0.2 \times 3 : 1.6) \times 3 = 0.6 : 4.8$

$P:O = 6.2 : 4.8 = 3.1 : 2.4 $ (ratio by mass) $= 1.29: 1$

$1.29$ mass $P= 1 $ mass $O$

$x$ mass $P= 16 \ g$ mass $O$

$x= 16 \times 1.29 = 20.64 \ g$ Phosphorous

$= \cfrac {20.64}{31}$ moles Phosphorous

$= 0.66$

$1 \ mole \ O(y) = 0.66 \ mole \ P(x) = \cfrac 23$

$\therefore \ x:y = 2:3 $

$x =2 \ ; \ y=3$

The enthalpies of combustion of carbon and carbon monoxide are -393.5 KJ and -283 KJ respectively the enthalpy of formation of carbon monoxide is :

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. -676.5 KJ

  2. -110.5 KJ

  3. 110.5 KJ

  4. 676.5 KJ

Show answer
Correct Option: C

The incorrect statement is :

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $ \Delta _{ So| }H^{ 0 }={ \Delta } _{ Lattice }H^{ 0+ }\Delta _{ Hyd }H^{ 0 } $

  2. The enthalpy on dilution is independent on original concentration.

  3. $ N \equiv N > C \equiv N > C \equiv C $ [value of mean bond enthalpy in KJ/mol]

  4. $ H _2O(S) \overset { 1\quad bar }{ \rightleftharpoons } H _2O(I) 273 K $


    $ \Delta U= +Ve; \Delta H = +ve; W= +ve; q = +ve $

Show answer
Correct Option: B

The enthalpy of tetramerization of $X$ in gas phase $(4X(g)\rightarrow { X } _{ 4 }(g))$ is $-100\ kJ/mol$ at $300\ K$. The enthalpy of vaporisation for liquid $X$ and ${X} _{4}$ are respectively $30\ kJ/mol$ and $72\ kJ/mol$ respectively.
$\Delta S$ for tetramerization of $X$ in liquid phase is $-125\ J/K mol$ at $300\ K$.
What is the $\Delta G$ at $300\ K$ for tetramerization of $X$ in liquid phase?

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $-52\ kJ/mol$

  2. $-98\ kJ/mol$

  3. $-14.5\ kJ/mol$

  4. $None\ of\ these$

Show answer
Correct Option: A

Heat of formation of $2$ moles of ${NH} _{3}(g)$ is $-90kJ$; bond energies of $H-H$ and$N-N$ bonds are $435kJ$ and $390kJ$ ${mol}^{-1}$ respectively. The value of the bond energy of $N\equiv N$ will be:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $-472.5\ kJ$

  2. $-945\ kJ$

  3. $472.5\ kJ$

  4. $945\ kJ$ ${mol}^{-1}$

Show answer
Correct Option: D
Explanation:

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

The standard enthalpy of formation of ${NH} _{3}$ is $-46kJ{ mol }^{ -1 }$. If the enthalpy of formation of ${H} _{2}$ from its atoms is $-436kJ{ mol }^{ -1 }$ and that of ${N} _{2}$ is $-712kJ{ mol }^{ -1 }$, the average bond enthalpy of $N-H$ bond in ${NH} _{3}$ is:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $+1056kJ{ mol }^{ -1 }\quad $

  2. $-1102kJ{ mol }^{ -1 }\quad $

  3. $-964kJ{ mol }^{ -1 }\quad $

  4. $+352kJ{ mol }^{ -1 }\quad $

Show answer
Correct Option: D
Explanation:

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

Calculate $ \Delta { H }^{ o }$ of the reaction:

$ { CH } _{ 2 }={ CH } _{ 2 }+3O=O\longrightarrow 2O=C=O+2H-O-H$

The average bond enthelpies of various bond are:

$Bond \quad \quad \quad\quad \quad C-H\quad \quad O=O\quad \quad C=O\quad \quad O-H\quad \quad C=C$
$Bond\ enthalpy$:     414               499               724           460               619
[kJ/mol]

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ

Show answer
Correct Option: C

$H _{2}+Cl _{2}\rightarrow 2HCl+44\ K.Cal$. Heat of decomposition of $HCl$ is:

chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

  1. $-44\ K.Cal$

  2. $+44\ K.Cal$

  3. $-22\ K.Cal$

  4. $+22\ K.Cal$

Show answer
Correct Option: D
Explanation:

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

Heat of decomposition of $HCl$ :-
$2HCl \longrightarrow \Delta H _{decomp}=+44 Kcal$
$1 HCl \longrightarrow \Delta H _{decomp}=+22 Kcal$
So, Heat of decomposition of $HCl=+22Kcal$