Chemistry Test-24

The equilibrium reaction is \(A ( g )+2 B ( g ) \rightleftharpoons C ( g )+ D ( g )\).

The initial moles of \(A , B , C\) and \(D\) are \(0.5,1,0.5\) and \(3.5\) moles respectively.

The equilibrium moles of \(A , B , C\) and \(D\) are \(0.5- x , 1-2 x , 0.5+ x\) and \(3.5+ x\) moles respectively.

The expression for the equilibrium constant is \(K _{ c }=\frac{[ C ][ D ]}{[ A ][ B ]^{2}}\).

Substitute values in the above expression.

\(K _{ c }=\frac{(0.5+ x )(3.5+ x )}{(0.5- x )(1-2 x )^{2}}\)

So, \(x=0.4999\) and \([B]=(1-2 x)=2 \times 10^{-4}\).

The activation energy in chemisorption is larger than that in physisorption. Consequently, an increase in temperature initially increases the rate of chemisorption as the heat supplied provides the activation energy for the process.

\(p \pi-p \pi\) back-bonding is the main factor responsible for the weak acidic nature of B-F bonds in BF₃.

The binding energy of a particle can be defined as the minimum energy required to remove nucleons (Proton or neutron) to an infinite distance from the nucleus. It can be expressed as:

\(\Delta E=\Delta m c^{2}\)

Whereas Binding energy per nucleus, \(E_{b n}=\frac{\Delta E}{A}\)

From the above, we can see that, as atom becomes heavier Binding energy increases. i.e., it means that it takes more energy to remove a single nucleus as atomic mass increase.

So, we can say that after fission when a heavier nucleus of atomic mass\(240\) breaks into lighter nucleus the new atom will have more binding energy and it will be more tightly bound and energy would be released in a form of heat and radiation.

Inhaling of nitrogen oxides by humans results in pulmonary edema and hemorrhage. Several sensitive plants get adversely affected by NOx even at the concentration of 1 ppm for a day or 0.35 ppm for a few months. The affected plants start shedding their leaves and fruit. Nitrogen oxides cause fading of textile dyes and deterioration of cotton and nylon fabrics.

During adsorption, there occurs an increase in entropy of the systemregarding adsorption is not true.

Given,

Molecular formula =Be_nAl₂Si₆O₁₈

The oxidation states of each element of given molecular formula:

Be = +2

Al = +3

Si = +4

O = -2

(2n) + (3 × 2) + (4 + 6) + (−2 × 18) = 0

⇒2n + 30 − 36 = 0

⇒2n = 6

⇒n = 3

Diphenyl amine is aromatic secondary amine. N atom is attached to two aryl groups and one H atom. Aniline is aromatic primary amine. Sec. butyl amine and tert. butyl amine are aliphatic primary amines.

Given :- \(E _n=\frac{-313.6}{n^2}\)

To calculate:- The value of \(n\) if the value of \(E _i=\) \(-40 k c a l\) mol \(^{-1}\)

\(E_i=-40\)

By the Nernst equation,

\(E_i = E_n\)

Therefore,

\(-40=\frac{-313.6}{n^2}\)

\(n^2=\frac{313.6}{40}\)

\(n^2=7.84\)

\(n= \sqrt 7.84\)

\(n=2.8 \approx 3\)

Energy is \(200 Mev\), converting into \(eV\) we get, \(200 \times 10^{6} ev\). But the standard SI unit is Joules, thus converting into \(200 \times 10^{6} \times 1.6 \times 10^{-19} J\). Also, the power is \(1000 kW\) converting it into watts we get \(1000 \times 1000=10^{6} W\)

We know power is given by \(P=\frac{n E}{t}\), putting the value we get,

\(\Rightarrow P=\frac{n E}{t}\)

\(\Rightarrow \frac{P}{E}=\frac{n}{t}\)

\(\Rightarrow \frac{n}{t}=\frac{10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\)

\(\therefore \frac{n}{t}=3.125 \times 10^{16}\)

So, the rate of nuclear fission comes out to be \(3.125 \times 10^{16}\).