Chemistry Test - 2

\(\frac{1}{2} N _{2( g )}+\frac{3}{2} H _{2( g )} \rightleftharpoons NH _{3( g )}\)
Partial pressure of \(NH _{3}=\frac{24}{100} \times 100\)
\(=24.0 atm\)
Pressure of \(N _{2}+ H _{2}=100-24=76 atm\)
Partial pressure of \(H _{2}=\frac{3}{4} \times 76=57 atm\)
Partial pressure of \(N _{2}\) \(=\frac{1}{4} \times 76=19 atm\)
\(K _{p}=\frac{ p _{ NH _{3}}}{ p _{ N _{2}}^{1/2}+ p _{ H _{2}}^{3/2}}\)
\(=\frac{24}{(19)^{1/2} \times(57)^{3/2}}\)
\(=\frac{24}{4.358 \times(7.549)^{3}}\)
\(=\frac{24}{4.358 \times 430}=0.0128 atm ^{-1}\)

III lonisation energy of \(L i^{+2}\) \(=\frac{Z^{2}}{n^{2}} E_{1} H^{\prime}\) where \((Z=3, n=1)\)
\(=-\frac{Z^{2} \times 2.18 \times 10^{-18}}{ n ^{2}} Jule\)
\(n =1\) for third ionisation.
\(=-\left(-9 \times 2.18 \times 10^{-18}\right) J\)
\(=19.62 \times 10^{-18} J\)

\(d U=d q+d W \ldots \ldots \ldots(i)\)
as process is isothermal \(d T=0\)
so \(d U=n R d T=0\)
so \(d q=-d W\)
now \(P V=n R T\)
for isothermal process \(P V=\) constant
\(\Rightarrow P d V+V d P=0\)
\(\Rightarrow P d V=-V d P \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(i i)\)
now \(d W=P d V\)
so, \(d q=-P d V\)
\begin{equation}\begin{aligned}
&\text { change in heat energy } \triangle q=\int d q=\int_{P_{1}}^{P_{2}}-P d V\\
&\Rightarrow \quad \Delta q=\int_{P_{1}}^{P_{2}} V d P\\
&\Delta q=\int_{P_{1}}^{P_{2}} \frac{n R T}{P} d P
\end{aligned}\end{equation}\begin{equation}\begin{array}{l}
\Delta q=n R T \int_{P_{1}}^{P_{2}} \frac{d P}{P}=n R T \ln \frac{P_{2}}{P_{1}}=-\frac{w}{M} R T \ln _{e} \frac{P_{1}}{P_{2}}\left(n=\frac{w \operatorname{eigh} t}{\text { Molecular weight }}\right) \\
\Delta q=-2.303 \frac{w}{M} R T \log _{10} \frac{P_{1}}{P_{2}} \\
\Delta q=-2.303 \frac{20}{4} \times 2 \times 300 \times \log _{10} \frac{2}{20} \\
=6909 cal =6.9 k \text { . cal. }
\end{array}\end{equation}

New carbon-carbon bond is formed is Friedel Craft reaction and Reimer-Tiemann reaction

Friedel Craft’s reaction

Reimer Tiemann Reaction

Hence option B is correct.

The reaction is an example of cine substitution or elimination addition.

For any gas \(22.4 L\) volume is occupied by \(=1\) mole
\(56 L\) will be occupied by \(=\frac{56}{22.4}=2.5 mole\) of ozone gas.
I molecule of Ozone \(O_{3}\) has 3 atoms of \(O\)
2.5mole \(O _{3}\) has
\(=2.5 \times 3=7.5\) mole of oxygen atoms
or \(7.5 N_{A}\) oxygen atoms

Equivalents of \(K M n O_{4}=\) equivalent of \(F e S O_{4}+\) equivalent of \(F e C_{2} O_{4}\)
\(X \times 5=1 \times 1+1 \times 3\)
\(X=\frac{4}{5}\) moles

\(H C l+N a O H \rightarrow N a C l+H_{2} O\)
In this chemical reaction, the molar ratio is 1: 1 between \(H C l\)
and \(NaOH.\)
So, moles of \(H C l=\) moles of
\(N a O H\)
\(M_{H C l} \times\) Volume of \(H C l=\)
\(M_{N a O H} \times\) Volume of \(N a O H\)
\(M_{H C l}=\frac{M_{N a O H} \times \text { volume of } N a O H}{\text { volume of } H C l}\)
\(M_{H C l}=\frac{25.00 ml \times 1.00 M}{50.00 ml }\)
\(M_{H C l}=0.50 M HCl\)
So, the concentration of \(H C l\) is \(0.50 M.\)

Balanced reaction is as follows:
\[P_{4} S_{3}+8 O_{2} \rightarrow P_{4} O_{10}+3 S O_{2}
\]Mass given: \(440 g \quad 384 g\)
No. of moles: \(\frac{440}{221} \frac{384}{32}\)
\[=1.99 \quad 12
\]Ratio of moles given and stoichiometry \(\quad \frac{1.99}{1} \quad \frac{12}{8}\)
coefficient:
\[=1.99 \quad 1.5
\]As 1.5 is lower than \(1.99 .\) So, \(O_{2}\) is limiting reagent and product will form according to number of moles of \(O_{2}\) (given).
So, Mole produced of \(P_{4} O_{10}=\frac{1}{8} \times 12=1.5 mol\)
And mass produced of \(P_{4} O_{10}=1.5 \times 284=426 g\).

In a closed container, \(N O_{2}\) was taken which dimerises to \(N_{2} O_{4}\) The reaction is,
\(2 N O_{2} \rightarrow N_{2} O_{4}\)
\(100-x \quad \frac{x}{2}\) where, \(x=\) number
of moles reacted
Mole fraction of \(N_{2} O_{4}=\frac{\frac{x}{2}}{(100-x)+\frac{x}{2}}=\frac{2}{3}\)
or, \(200-2 x+x=\frac{3 x}{2}\)
or, \(200=1.5 x+x\)
\(x=\frac{200}{2.5}=80\)
When divided by 10 , we get \(\frac{80}{10}=8.\)