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Chemistry 3 Marks Question

Model Paper 2 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 7 questions.

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Question 13 Marks
Describe what you need to do in the laboratory to test
i. the law of conservation of mass,
ii. the law of definite proportion
iii. the law of multiple proportions.
Answer
i. In order to test the law of conservation of mass, a reaction would have to be carried out in which the mass of the reactants and the mass of the products are weighed and shown to be the same.
ii. The law of definite proportions could be shown by demonstrating that no matter, how a compound is obtained, the reactants remain at the same proportions by mass. This can be done by decomposing a compound and showing that the masses of the elements present are always in the same whole number ratio.
iii. To test the law of multiple proportions, two different compounds made up of the same elements combining with that of the elements in different samples would have to be in the small whole number ratio.
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Question 23 Marks
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer
Electronic configurations of Na and Mg are
$
\begin{aligned}
& Na=1 s^2 2 s^2 2 p^6 3 s^1 \\
& Mg=1 s^2 2 s^2 2 p^6 3 s^2
\end{aligned}
$
The $1^{\text {st }}$ ionization enthalpy of magnesium is higher than that of Na due to higher nuclear charge and slightly smaller atomic radius of Mg than Na.
Electronic configurations of Na and Mg after loosing 1 electron are
$
\begin{aligned}
& Na^{+}=1 s^2 2 s^2 2 p^6 \\
& Mg^{+}=1 s^2 2 s^2 2 p^6 3 s^1
\end{aligned}
$
After the loss of the first electron, $Na ^{+}$formed has the electronic configuration of neon $(2,8)$. The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand, $Mg ^{+}$ formed after losing the first electron still has one more electron in its outermost orbital. Therefore, the second ionization enthalpy of magnesium is much smaller than that of sodium.
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Question 33 Marks
A photon of wavelength $4 \times 10^{-7} m$ strikes on the metal surface, the work function of metal being 2.13 eV . Calculate
i. the energy of the photon $( eV )$,
ii. the kinetic energy of emission,
iii. the velocity of the photoelectron. $\left(1 eV =1.6020 \times 10^{-19} J\right)$.
Answer
i. The energy of the photon
$
\begin{aligned}
& \text { Energy }(E)=\frac{hc}{\lambda}-\frac{\left(6.626 \times 10^{-34} Js\right) \times\left(3 \times 10^8 ms^{-1}\right)}{\left(4 \times 10^{-7} m\right)}=4.97 \times 10^{-19} \\
& =\frac{(1 eV)}{\left(1.602 \times 10^{-19} J\right)} \times\left(4.97 \times 10^{-19} J\right)=3.1 eV
\end{aligned}
$
ii. The kinetic energy of emission
Kinetic energy of emission $= E$ - work function (i.e. kinetic energy of emitted electron) $=(3.1-2.13)=0.97 eV$
iii. Velocity of photoelectron
$
\begin{aligned}
& \text { KE of emission }=\frac{1}{2} mv^2=0.97 eV \\
& =0.97 \times 1.602 \times 10^{-19} J=0.97 \times 1.602 \times 10^{-19} kg m^2 s^{-2} \\
& \text { or } v^2=\frac{2 \times 0.97 \times 1.602 \times 10^{-19}\left(kgm^2 s^{-2}\right)}{\left(9.1 \times 10^{-31} kg\right)}=0.34 \times 10^{12} m^2 s^{-2} \\
& \text { or } v=\left(0.34 \times 10^{12} m^2 s^{-2}\right)^{1 / 2}=0.583 \times 10^6 ms^{-1}=5.83 \times 10^5 ms^{-1}
\end{aligned}
$
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Question 43 Marks
Justify that the following reactions are redox reactions:
i. $CuO + H _2(g) \longrightarrow Cu ( s )+ H _2 O$ (g)
ii. $Fe _2 O _3$ (s) $+3 CO ( g ) \longrightarrow 2 Fe ( s )+3 CO _2$ (g)
iii. $4 BCl _3(g)+3 LiAlH _4(s) \longrightarrow 2 B_2 H _6(g)+3 LiCl ( s )+3 AlCl _3(s)$
Answer
The given equations for different reactions are :
Image
As per above equation, it is noted that.
a. an atom of oxygen $( O )$ is removed from CuO ,
$\therefore$ it is reduced to Cu , while
b. O is added to $H _2$ to form $H _2 O$
$\therefore$ it is oxidized.
Further,
Oxidation number. of Cu decreases from +2 in CuO to 0 in Cu , 
oxidation number of H increases from 0 in $H _2$ to +1 in $H _2 O$.
$\therefore CuO$ is reduced to Cu but $H _2$ is oxidized to $H _2 O$.
Thus, the reaction is a redox reaction.
Image
In the above equation for reaction, it is seen that
a. The oxidation number of Fe decreases from +3 in $Fe _2 O _3$ to 0 in Fe , and oxidation number of C increases from +2 in CO to +4 in $CO _2$.
Further, 
oxygen is removed from $Fe _2 O _3$, and added to CO to form $CO _2$ 
therefore, $Fe _2 O _3$ is reduced while CO is oxidized. 
Thus, the given reaction is a redox reaction.
Image
Oxidation number of B decreases from +3 in $BCl _3$ to -3 in $B _2 H _6$ 
while,
oxidation number of H increases from -1 in $LiAlH _4$ to +1 in $B _2 H _6$. 
Therefore, $BCl _3$ is reduced and
$LiAlH _4$ is oxidized.
Further, it is noted that, 
$H$ is added to $B$ forming $B_2 H6$ from $BCl _3$ but is removed from $LiAlH _4$, therefore,
$BCl _3$ is reduced while $LiAlH _4$ is oxidised.
Thus, 
the redox nature of above reaction is justfied.
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Question 53 Marks
Give the relationship between $\Delta U$ and $\Delta H$ for gases.
Answer
Let $V_A$ be the total volume of gaseous reactants,
$V _{ B }$ be the total volume of gaseous product.
Let $n _{ A }$ be the number of moles of the reactant,
$n _{ B }$ be the number of moles of the product,
At constant pressure and temperature,
$
\begin{aligned}
& p V_A=n_A R T \\
& p V_B=n_B R T \\
& \Rightarrow pV_{B}-pV_{A}=\left(n_{B}-n_{A}\right) RT \\
& \Rightarrow p \Delta V=(\Delta n)_g RT
\end{aligned}
$
Here, $(\Delta n)_g=n_B-n_A$ is equal to the difference between the number of moles of gaseous products and gaseous reactants. We know that,
$
\Delta H=\Delta U+(\Delta n)_g R T
$
Now, $\Delta H = q _{ p }$ (heat change under constant pressure),
$\Delta U = q _{ v }$ (heat change under constant volume).
Therefore, $q_p=q_v+(\Delta n)_g R T$
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Question 63 Marks
1. At $298 K, K _{ p }$ for the reaction $N _2 O _4(g) \rightleftharpoons 2 NO _2(g)$ is 0.98 . Predict whether the reaction is spontaneous or not.
2. Define specific heat.
3. State Hess's law.
Answer
1. Given, At $298 K, N _2 O _4(g) \rightleftharpoons 2 NO _2(g)$.
We know that, $\Delta_r G^{\circ}=-2.303 R T \log K_p$
According to the question, $K _{ p }=0.98$.
As, $K _{ p }<1, \Delta_r G^{\circ}$ will be positive. $[\because \log (0.98)=-0.0087]$
Hence, the reaction is non-spontaneous.
2. Specific heat: The quantity of heat required to raise the temperature by one degree Celsius (or one Kelvin) of one unit mass of a substance is known as specific heat.
3.The change of enthalpy of a reaction remains same whether the reaction is carried out in one step or several steps.
$
\Delta H=\Delta H_1+\Delta H_2+\Delta H_3 \ldots
$
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Question 73 Marks
Explain the structure of the $CO _2$ molecule.
Answer
The experimentally determined carbon to oxygen bond length in $CO _2$ is 115 pm . The lengths of a normal carbon to oxygen double bond $( C = O )$ and carbon to oxygen triple bond $( C \equiv O )$ are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in $CO _2(115 pm )$ lie between the values for $C = O$ and $C \equiv O$. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of $CO _2$ is best described as a hybrid of the canonical or resonance forms I, II and III.
Image
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