Tag: chemistry

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Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For the reaction, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, if the concentration of $N{ O } _{ 2 }$ increases by $5.2\times { 10 }^{ -3 }M$ in $100$ sec, then the rate of reaction is:

  1. $1.3\times { 10 }^{ -5 }M{ s }^{ -1 }$
  2. $5\times { 10 }^{ -4 }M{ s }^{ -1 }$
  3. $7.6\times { 10 }^{ -4 }M{ s }^{ -1 }$
  4. $2\times { 10 }^{ -3 }M{ s }^{ -1 }$
  5. $2.5\times { 10 }^{ -5 }M{ s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vartheta =\cfrac { 1 }{ 4 } \cfrac { d }{ dt } [{ NO } _{ 2 }]=\cfrac { 1 }{ 4 } (5.2\times { 10 }^{ -5 }M{ s }^{ -1 })\ =1.3\times { 10 }^{ -5 }{ Ms }^{ -1 }$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

A reaction is represented as $2A + B \mapsto  2C + 3D$. The concentration of C at 10 s is 4 moles $l^{-1}$. The concentration of C at 20 seconds is 5.2 moles $l^{-1}$. The rate of reaction of B in the same time interval could be :

  1. $-0.12$ mole $l^{-1} S^{-1}$
  2. $-0.6$ mole $l^{-1} S^{-1}$
  3. $-0.06$ mole $l^{-1} S^{-1}$
  4. $-1.2$ mole $l^{-1} S^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$r _{B} = \dfrac{1}{2}\dfrac{\Delta C}{\Delta t}= \dfrac{-1}{2}\dfrac{(1.2)}{10}$


$(\Delta C= 5.2 -4 = 1.2, \Delta t= 10 sec.)$

$= -0.06$ mole l$^{-1}$ sec$^{-1}$
Hence the answer is $C$.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For $S{O _2}C{l _{2\left( g \right)}} \to S{O _{2\left( g \right)}} + C{l _{2\left( g \right)}},$ Pressures of $S{O _2}C{l _2}$ at $t = 0$ and $t = 20$ minutes respectively are $700mm$ and $350mm.$ When $\log \left( {{P _0}/p} \right)$ is plotted against time ($t$), slope equals to:

  1. $1.505 \times {10^{ - 2}}{s^{ - 1}}$
  2. $1.202 \times {10^{ - 3}}{\min ^{ - 1}}$
  3. $1.505 \times {10^{ - 2}}{\min ^{ - 1}}$
  4. $0.3465\ {\min ^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$k=\dfrac{2.303}{t}log\dfrac{P _0}{P}$ [1st order reaction]

Plot of $log\dfrac{P _0}{P}$ against $t$ will give slope $=  \dfrac{k}{2.303}$
Using given data, $k=\dfrac{2.303}{20}log \dfrac{700}{350}$
$\dfrac{k}{2.303}=\dfrac{log2}{20}=1.5\times 10^{-2}s^{-1}$
Slope $=1.5\times 10^{-2}s^{-1}$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

From the concentrations of R at different times given below. Determine  the average rate of the reaction range: R $\rightarrow$ P in given intervals of time.

t (s) 0 5 10 20 30
$10^{-3}\, \times\, [R] \,(mol\, L^{-1})$ 160 80 40 10 2.5
  1. $3.5\times\,10^{2}$ to $0.42 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
  2. $7\times\,10^{2}$ to $0.84 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
  3. $8\times\,10^{3}$ to $0.37 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$
  4. $16\times\,10^{3}$ to $0.75 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average Rate $=\cfrac{ change \ in \ concentration }{ change \ in \ time }$


$\therefore$ here we wanted to find reaction range

Initially avergae rate$=\cfrac{ (160-80) }{ (5-0) }=$$\cfrac{ 80 }{ 5 }$
$=16$

Here, ${ 10 }^{ 3 }$ already given.

$\therefore$ Average rate $=16 \times {10}^{3}\ mol {L}^{-1} {s}^{-1}$

Final average rate $=\cfrac{ (10-2.5) }{(30-20) }=\cfrac{ 7.5 }{ 10 }=0.75$

$\therefore$ Average Rate $=0.75 \times {10}^{3}\ mol {L}^{-1} {s}^{-1}$

Multiple choice chemistry grouping of materials properties of materials grouping and classification of objects nature of things

During the formation of rain, when water vapour changes back to liquid in the form of rain drops:

  1. Heat is absorbed

  2. Heat is released

  3. Heat is first absorbed and then released

  4. There is no exchange of heat
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The energy contained in the gaseous form (vapour) is more than the energy contained in the liquid form (water). Therefore when vapour changes to water it loses some heat.

Multiple choice chemistry atoms, molecules and radicals history of atomic model atomic theory of dalton daltons atomic theory

______ are the smallest unit of matter that can take part in a chemical reaction.

  1. Molecules

  2. Compounds

  3. Atoms

  4. Electrons
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Dalton's atomic theory states that atoms are the smallest unit of matter that can participate in a chemical reaction. Chemical reactions involve the combination, separation, or rearrangement of atoms, but atoms are neither created, destroyed, divided into parts, or converted into atoms of any other kind.