3 Marks Question

Question 13 Marks

1. Give the mathematical expression of enthalpy.
2. Neither q nor W is a state function but $q + W$ is a state function. Explain why?
3. The standard heat of formation of $Fe _2 O _3$ (s) is $824.2 kJ mol ^{-1}$ Calculate heat change for the reaction.
$
4 Fe(s)+3 O_2(g) \rightarrow 2 Fe_2 O_3(s)
$

Answer

1. Mathematical expression of enthalpy is $H = U + pv$, where U is internal energy.
2. q and W are not state functions. But as we know that, $q + W =\Delta U$, which is a state function.
Hence, $q + W$ is a state function.
3. According to the question, the standard heat of formation of $Fe _2 O _3$ (s) is $824.2 kJ mol ^{-1}$.
The standard heat of formation of Fe and $O _2$ is zero because they are in their basic standard states.
Reaction:
$
4 Fe(s)+3 O_2(g) \rightarrow 2 Fe_2 O_3(s)
$
We know that, $\Delta H^{\circ}=\sum \Delta H_f^{\circ}$ (products $)-\sum \Delta H_f^0$ (reactants)
$
\begin{aligned}
& =\left[2 \times \Delta H_{f}^0 Fe_2 O_3(s)\right]-\left[4 \Delta H_{f}^0 Fe(s)+3 \Delta H_{f}^0 O_2(g)\right] \\
& =2(-824.2)-[4 \times 0+3 \times 0] \\
& =-1648.4 kJ
\end{aligned}
$

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Question 23 Marks

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2 A+4 B \rightarrow 3 C+4 D$ , when 5 moles of A react with 6 moles of $B$, then
i. Which is the limiting reagent?
ii. Calculate the amount of C formed?

Answer

The given equation is : $2 A+4 B \rightarrow 3 C +4 D$
i. It is clear from the above equation that: 2 moles of ' $A$ ' requires 4 moles of ' $B$ ' for the reaction i.e. ratio of moles of $A$ to $B$ is 2: 4 or $1: 2$. Hence, for 5 moles of ' $A$ ', the moles of ' $B$ ' required $=5 mole$ of $A \times \frac{4 mol \text { of } B}{2 mol \text { of } A}=10 mol$ of $B$. But we have only 6 moles of ' $B$ ', hence, ' $B$ ' is the limiting reagent.
ii. Since 4 moles of ' $B$ ' gives 3 moles of ' $C$ '. Hence, 6 moles of ' $B$ ' will produce $\frac{3}{4} \times 6=4.5$ mole of $C$.

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Question 33 Marks

The electronic configuration of some elements are given below:
a. $1 s^2, 2 s^2, 2 p^6, 3 s^2$
b. $1 s^2, 2 s^2, 2 p^6$
c. $1 s^2, 2 s^2, 2 p ^2$
d. $1 s^2, 2 s^2, 2 p ^6, 3 s^1$
e. $1 s^2, 2 s^2, 2 p^5$
Answer the following questions:
i. Name the elements.
ii. Which of these have the lowest Ionization enthalpy?
iii. Which is a halogen?
iv. Which is an alkali metal?
v. Which is an inert gas?

Answer

i. (a) Magnesium, (b) Neon (c) Carbon (d) Sodium (e) Flourine
ii. $1 s^2, 2 s^2, 2 p ^6, 3 s^1$ (Sodium)
iii. $1 s^2, 2 s^2, 2 p ^5$ (Flourine)
iv. $1 s^2, 2 s^2, 2 p^6, 3 s^1$ (Sodium)
v. $1 s^2, 2 s^2, 2 p^6($ Neon $)$

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Question 43 Marks

What transition in a hydrogen spectrum would have the same wavelength Balmer transition $n =4$ to $n =2$ of $\overline{ v }=\frac{1}{\lambda}= R _{ H } Z^2\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$ spectrum?

Answer

For an atom, $\overline{ v }=\frac{1}{\lambda}= R _{ H } Z^2\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$
For $He ^{+}$spectrum $Z =4, n _2=4, n _1=2$
$\therefore$ For hydrogen spectrum: $\overline{ v }=\frac{3 R _{ H }}{4}$ and $Z =1$
$\therefore \overline{ v }=\frac{1}{\lambda}= R _{ H } \times 1\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$
or $R_H\left(\frac{1}{n_1^2}-\frac{1}{m_2^2}\right)=\frac{3 R_H}{4}$ or $\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}$
This corresponding to $n _1=1, n _2=2$ and means that the transition has taken Lyman series from $n =2$ to $n =1$.
Thus, the transition is from $n _2$ to $n _1$ in case of hydrogen spectrum.

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Question 53 Marks

Identify the type of redox reaction taking place in the following.

Answer

i. Combination reaction
ii. Displacement reaction
iii. Decomposition reaction
iv. Metal displacement reaction
v. Non-metal displacement reaction
vi. Disproportionation reaction

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Question 63 Marks

100 mL of a liquid is contained in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1 mL at this constant pressure. Find $\Delta H$ and $\Delta U$.

Answer

According to the question, $p_1=1$ bar, $p_2=100 bar , V_1=100 mL, V_2=99 mL$.
We know that, $\Delta U=q+W$
For the adiabatic process, $q =0$
So, $\Delta U=W$
Now, $W=-p \Delta V=-100(99-100)=100$ bar ml
We know that, $\Delta H=\Delta U+\Delta p V$
$
\begin{aligned}
& =100+p_2 V_2-p_1 V_1 \\
& =100+(100 \times 99)-(1 \times 100) \\
& =100+9900-100 \\
& =9900 \text { bar } mL
\end{aligned}
$

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Question 73 Marks

Describe the change in hybridisation (if any) of the Al atom in the following reaction.
$
AlCl_3+Cl^{-} \rightarrow AlCl_4^{-}
$

Answer

The valence orbital picture of aluminum in the ground state can be shown as:

Hence, it undergoes $sp ^2$ hybridization to give a trigonal planar arrangement (in $AlCl _3$ ). To form $AlCl _4^{-}$, the empty 3 pz orbital also gets involved and the hybridization changes from $sp ^2$ to $sp ^3$. As a result, the shape becomes tetrahedral.

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Chemistry 3 Marks Question

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