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

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

  1. $H _{2}S$ and $SO _{2}$
  2. $NH _{3}$ and $NO _{2}$
  3. $N _{2}O$ and NO
  4. $N _{2}S$ and $Na _{2}O$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $2:1$

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

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $1:2$.

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

$P _xO _y$ 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 $0.2 : 1.6$?

  1. $1$ and $1$
  2. $1$ and $2$
  3. $2$ and $3$
  4. $2$ and $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

$x =2 \ ; \ y=3$

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

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 :

  1. -676.5 KJ

  2. -110.5 KJ

  3. 110.5 KJ

  4. 676.5 KJ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The enthalpy of formation of carbon monoxide is calculated using Hess's law by subtracting the enthalpy of combustion of carbon monoxide from that of carbon. Specifically, delta H_f(CO) = delta H_c(C) - delta H_c(CO) = -393.5 - (-283) = -110.5 kJ. This represents the energy change when one mole of carbon monoxide is formed from its elements.

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

The incorrect statement is :

  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 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The statement about dilution enthalpy is incorrect because enthalpy of dilution DOES depend on the original concentration of the solution. Dilution from different starting concentrations releases different amounts of heat. The other statements are correct: Born-Haber cycle relation, bond enthalpy order (N≡N=946 kJ/mol > C≡N=891 > C≡C=619), and ice melting thermodynamics.

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

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?

  1. $-52\ kJ/mol$
  2. $-98\ kJ/mol$
  3. $-14.5\ kJ/mol$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

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:

  1. $-472.5\ kJ$
  2. $-945\ kJ$
  3. $472.5\ kJ$
  4. $945\ kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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:

  1. $+1056kJ{ mol }^{ -1 }\quad $
  2. $-1102kJ{ mol }^{ -1 }\quad $
  3. $-964kJ{ mol }^{ -1 }\quad $
  4. $+352kJ{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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]

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using bond enthalpy method: ΔH = Σ(bonds broken) - Σ(bonds formed). Breaking: 4 C-H (4×414=1656), 1 C=C (619), 3 O=O (3×499=1497). Total broken = 3772. Forming: 4 C=O (4×724=2896), 4 O-H (4×460=1840). Total formed = 4736. ΔH = 3772-4736 = -964 kJ. The reaction is exothermic.

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

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

  1. $-44\ K.Cal$
  2. $+44\ K.Cal$
  3. $-22\ K.Cal$
  4. $+22\ K.Cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $HCl=+22Kcal$