Chemistry Test-3

The adsorption of gases on solid surface is exothermic because the entropy decreases.

\(\mathrm{Mg}\) metal evolves hydrogen on reacting with cold dilute \(\mathrm{HNO}_3\).Only magnesium and manganese react with nitric acid to release hydrogen gas.

Hydrogen gas is not evolved when other metals react with nitric acid because it is a strong oxidising agent. It oxidises hydrogen produced to water and itself gets reduced to a nitrogen oxide.

Benzene is a polymer of Ethyne.

\(C ( s )+ CO _{2}(g) \rightleftharpoons 2 CO (g)\)

\(K_{P_{1}}^{\prime}=10^{-12} atm ^{-1} \quad \ldots\) (i)

\(\therefore K_{P_{1}}^{\prime}=\frac{1}{K_{p_{1}}}=\frac{1}{10^{-12}} atm\)

\(2 CO (g)+2 Cl _{2}(g) \rightleftharpoons 2 COCl _{2}(g) ; K_{p_{2}}^{\prime} \ldots \text { (ii) }\)

\(K_{p^{\prime}}=\left(K_{p_{2}}\right)^{2}=\left(3 \times 10^{-3}\right)^{2} atm ^{-2}\)

Adding Eqs. (i) and (ii), we get

\(C ( s )+ CO _{2}(g)+2 Cl _{2}(g) \rightleftharpoons2COCl_{2}(g)\)

\(K_{p}=10^{12} \times 9 \times 10^{-6}=9 \times 10^{6}\)

\(\Delta r_{g}=n_{p}-n_{r}(g)\)

\(\Delta r_{g}=2-(1+2)=-1\)

\(K_{p}=K_{0}(R T)^{-1}\)

\(K_{c} =K_{\rho}(R T)\)

\(K_{c} =9 \times 10^{6} \times 0.0821 \times 1140\)

\(=8.42 \times 10^{8} M ^{-1}\)

Primary valencies are also known as the oxidation state of the central metal atom.

\(K _{2}\left[ N i ( C N )_{4}\right]\)

Let \(x\) be the oxidation state of \(Ni\).

Now,

\(2 \times 1+x+4 \times(-1)=0\)

\(\Rightarrow x=+2\)

In Rutherford's alpha-particle scattering experiment, Very thin sheets of gold foil were bombarded with fast-moving alpha particles.

Given,

The percentage composition of hydrogen is 4.07%.

The percentage composition of carbon is 24.27%.

The percentage composition of oxygen is 17.65%.

We can write the steps for determining the empirical formula of a compound as follows:

The molar mass of hydrogen is \(1 \mathrm{~g} / \mathrm{mol}\).

The molar mass of carbon is \(12 \mathrm{~g} / \mathrm{mol}\).

The molar mass of oxygen is \(35.5 \mathrm{~g} / \mathrm{mol}\).

The moles of the element are calculated using the mass divided by their molar mass. We can write the expression to calculate of moles of an element as,

Moles of element \(=\frac{\text { Mass }}{\text { Molecular mass }}\)

We can calculate the moles of each element now,

The moles of hydrogen atom \(=\frac{4.07 g}{1 \mathrm{~g} / \mathrm{mol}}=4.07\) moles

The moles of carbon atom \(=\frac{24.27 \mathrm{~g}}{12 \mathrm{~g} / \mathrm{mol}}=2.02 \) moles

The moles of oxygen atom \(=\frac{71.65 g}{35.5 g / m o l}=2.01\) moles

The moles of hydrogen, carbon and oxygen are \(4.07\) moles, \(2.02\) moles and\(2.01\) moles respectively.

Let us now divide all values with least value obtained,

\(H=\frac{4.07}{2.01}=2\)

\(C=\frac{2.02}{2.01}=1\)

\(C l=\frac{2.01}{2.01}=1\)

We can write the empirical formula of the compound as \(C H_{2} C l\).

At equilibrium, Gibbs free energy change \(\left(\Delta \mathrm{G}^{\circ}\right)\) is equal to zero. The following thermodynamic relation is used to show the relation of \((\Delta \mathrm{G}^{\circ})\) with the enthalpy change \(\left(\Delta \mathrm{H}^{\circ}\right)\) and entropy change \(\left(\Delta \mathrm{S}^{\circ}\right)\).

\(\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{o}-\Delta \mathrm{S}^{o}\)

\(0=30 \times 10^{3}\left(\mathrm{~J} \mathrm{~mol}^{-1}\right)-\mathrm{T} \times 105\left(\mathrm{~J} \mathrm{~K}^{-1}\right) \mathrm{mol}^{-1}\)

Therefore,

\(T=\frac{30 \times 10^{3}}{105} \mathrm{~K}=285.71\mathrm{~K}\)

No molecule can have a zero order, i.e. a molecule with zero order cannot exist. On calculating the bond order of the given molecules, we get,

(b) \(\mathrm{Be}_{2}(4+4=8)=\sigma 1 \mathrm{~s}^{2},{ }^{*} 1 \mathrm{~s}^{2}+\frac{*}{\sigma} 2 \mathrm{~s}^{2}\)

\(\mathrm{BO}=\frac{4-4}{2}=0\)

(a) \(\mathrm{Be}^{+} 2(4+4-1=7)=\sigma 1 \mathrm{~s}^{2} \cdot{ }^{*} 1 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2} \cdot{ }^{*} 2 \mathrm{~s}^{1}\)

\(\mathrm{BO}=\frac{4-3}{2}=0.5\)

(c) \(B_{2}(5+5=10)\)

\(=\sigma 1 s^{2} \cdot{ }^{*} 1 s^{2}, \sigma 2 s^{2}, \pi 2 p_{y}^{1}\)

\(B O=\frac{6-4}{2}=1\)

(d) \(L i_{2}(3+3=6)\)

\(\sigma 1 s^{2}, \sigma 1 s^{2}, \sigma 2 s^{2}\)

\(B O=\frac{4-2}{2}=1\)

As mentioned, no molecule can have a zero order, it does not exist.