Chemistry Test - 14

We have to find the equivalent conductivity of a solution, which can be called the conductivity of a substance per gram

The equivalent of the same substance. This can be written as:

Equivalent conductivity \(=\frac{\text { conductivity }}{\text { gram equivalent }}\)

Now, we know that one gram equivalent of a substance is the number of gram equivalents per unit volume.

Gram equivalents \(=\frac{\text { No. of gram equivalents }}{\text { Volume }}\)

Here, we will consider the volume to be in \(m^{3}\) as it is the same as the volume in liters which is regularly considered to

Calculate the gram equivalents. Thus, taking the units for the gram equivalent:

Gram equivalents \(=\frac{g \text { equiv }}{m^{3}}\)

Substituting these units in the formula for equivalent conductivity we get,

Equivalent conductivity \(=\frac{\frac{S}{m}}{\frac{g \text { equiv }}{m^{3}}}\)

Now solving and simplifying, we get,

Equivalent conductivity \(=\frac{S}{m} \times \frac{m^{3}}{g \text { equiv }}\)

Equivalent conductivity \(=\frac{S m^{2}}{g \text { equiv }}\)

Equivalent conductance \(=\mathrm{cm}^{2} gm\) equiv \(^{-1}\)

The first law states that if \(\Delta Q\) is the amount of heat supplied to a thermodynamical system, some fraction of the heat energy is used to change the internal energy of the system andto do work.

\(\Delta Q=\Delta U+\Delta W\)

Where \(\Delta Q=\) the amount of heat supplied, \(\Delta U=\) Change in the internal energy \(\Delta W=\) work done.

• Indicator diagrams are plotted pressure against volume in a thermodynamic process.
• The area under the indicator diagram/P-V diagram shows total work done in a thermodynamical process.

According to the first law of thermodynamics

\(\Delta Q=\Delta U+\Delta W\)......(i)

In a cyclic process, the change in internal energy is zero i.e., \(\Delta U=0\)

\(\Delta Q=\Delta W\)........(ii)

The area under the P-V diagram will give you work done

The area of a rectangle \(=\) Length \(×\) Breadth \(=A B \times A D\)

\({AB}=3 {~V}-{V}=2 {~V}\)

\({AD}=2 {P}-{P}={P}\)

Then the area of rectangle \(=2 {~V} {\times P}=2 {PV}\)

\(W=2 P V\)

Therefore, the work done \({W}=2 {PV}\)

Third law of thermodynamics states that the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero.

Mathematically, the absolute entropy of any system at zero temperature is the natural log of the number of ground states times Boltzmann's constant. The entropy of a perfect crystal lattice iszero, provided that its ground state is unique, because ln(1) = 0. So the option (A) is incorrect.

The entropy of a perfect crystal at absolute zero is exactly equal to zero. At absolute zero (zero kelvin), the system must be in a state with the minimum possible energy, and the above statement of the third law holds true provided that the perfect crystal has only one minimum energy state.

So according to the above statement at absolute zero the entropy of a perfectly crystalline substance is zero. So option (B) is correct.

Option (C) is incorrect as written above that at 0 kelvin the entropy of a perfectly crystalline substance is taken zero so it cannot be at 0 degree Celsius.Similarly option (D) is incorrect.

Nucleophiles are the species which have an excess of electrons.

Among the given species, the lone pair of nitrogen of pyrrole (B) is involved in the delocalization of the ring, thus, they are not available for donation. 

In aniline (D), the lone pair is involved in conjugation with the π electrons of the ring while in pyridine, these are relatively free for donation. Thus, the nitrogen of pyridine (A) is the most nucleophilic.

Phenyl and −COCH₃ ​both are electron-withdrawing groups, thus decreases the nucleophilicity of nitrogen.

\(\ln \frac{k_{310}}{k_{300}}=\frac{E_{a}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)\)

\(\ln \frac{k_{310}}{k_{300}}=\frac{E_{a}}{R}\left(\frac{310-300}{300 \times 310}\right)\)

Thus, when, the value of activation energy is maximum, the ratio \(\frac{\mathrm{k}_{310}}{\mathrm{k}_{300}}\) will be maximum.

The reaction \(\mathrm{X}+\mathrm{Y} \rightarrow \mathrm{Z} ; \mathrm{E}_{\mathrm{a}}=100 \mathrm{~kJ}\) has the maximum value of the activation energy.

Given:

Volume of bulb A \(=100\) ml

After opening the stopcock, the pressure for bulb A and B will become same i.e., \(0.40 ~P_{A}\) and the net volume will be \(V_{A}+V_{B}\).

Applying Boyle's Law,

\(P_{1} V_{1}=P_{2} V_{2}\)

Or \(P_{A} V_{A}=\left(0.40 P_{A}\right)\left(V_{A}+V_{B}\right)\)

Or \(P_{A} \times 100=\left(0.40 P_{A}\right)\left(100+V_{B}\right)\)

\(\Rightarrow V_{B}=150 \mathrm{ml}\)

\(\therefore\)The volume of B must be 150 ml.

We know,

Activation energy is given by:

\(k=A e^{\frac{-E_{a} }{R T}}\)

Case I:

\(k_{1}=A e^{\frac{-100 }{R T}}, k_{2}=A e^{\frac{-25 }{R T}}\)

\(\frac{k_{1}}{k_{2}}=\frac{e^{\frac{-100 }{R T}}}{e^{\frac{-25 }{ R T}}}=e^{\frac{-75} {R T}}\)

log. \(\frac{k_{2}}{k_{1}}=\frac{-75} {R T}=\frac{75 \times 10^{3}}{8.314 \times 293}\)

\(=30.788\)

\(\frac{k_{2}}{k_{1}}=2.35 \times 10^{30}\)

Case II:

\(r=k[A]^{n}\)

\(n\) and \([A]\) are same for cases I andII

\(\therefore \frac{r_{2}}{r_{1}}=\frac{k_{2}}{k_{1}}=2.35 \times 10^{30}\)

Therefore by comparison the values they remains unaffected, so does the reaction.

Structure of \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\) is shown in figure. Oxidation state of \(\mathrm{P}\) is +5 and it has one \(\mathrm{P}-\mathrm{O}-\) P bond.

We know that,

\(L_{n}=I \omega_{n}\)

\(L_{n}=\frac{n h}{2 \pi}\)

\(E=\frac{1}{2} I \omega_{n}{ }^{2}\)

Let \(\omega_{n}\) be the angular velocity of the molecule in \(n^{t h}\) level

Angular momentum of an orbiting electron in a molecule in \(n^{\text {th }}\) level is given by,

\(L_{n}=I \omega_{n}\) .............(1)

Where, I is the rotational inertia

From Bohr's postulate we know,

\(L_{n}=\frac{n h}{2 \pi}\) .........(2)

Where, \({n}\) is the principal quantum number of the molecule

From the equation. (1) and equation. (2) we get,

\(I \omega_{n}=\frac{n h}{2 \pi}\)

Rearranging above equation we get,

\(\omega_{n}=\frac{n h}{2 \pi I}\) ...............(3)

Rotational energy of a molecule in \(n^{\text {th }}\) level is given by,

\(E=\frac{1}{2} I \omega_{n}{ }^{2}\)

Substituting equation. (3) in above equation we get,

\(E=\frac{1}{2} I \frac{n h^{2}}{2 \pi I}\)

\(\Rightarrow E=\frac{1}{2} I \frac{n^{2} h^{2}}{4 \pi^{2} I^{2}}\)

\(\Rightarrow E=\frac{n^{2} h^{2}}{8 \pi^{2} I}\)

Rotational energy in the \(n^{t h}\) level \(\left(\mathrm{n}=0\right.\) is not allowed) is \(n^{2}\left(\frac{h^{2}}{8 \pi^{2} I}\right)\).

Green house gases that are generally responsible for global warming, are carbon dioxide and methane.

A greenhouse gas is any gaseous compound in the atmosphere that is capable of absorbing infrared radiation, thereby trapping and holding heat in the atmosphere. By increasing the heat in the atmosphere, greenhouse gases are responsible for the greenhouse effect, which ultimately leads to global warming.

Some green house gases are water vapour \((\mathrm{H}_{2} \mathrm{O})\), carbon dioxide \((\mathrm{CO}_{2})\), methane \((\mathrm{CH}_{4})\), nitrous oxide \((\mathrm{N}_{2} \mathrm{O})\), ozone \((\mathrm{O}_{3})\) and some artificial chemicals such as chlorofluorocarbons \(\mathrm{CFC}_{\mathrm{s}}\) and Hydrofluoric carbons.

Human activities increases concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{CH}_{4}\) in the atmosphere since the beginning of the industrial revolution.