TS EAMCET 2018 :: Chemistry :: General questions

• Chemistry - General questions

 If equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{0} C$  , standard free energy change of the process is 

 $(R= 8.314 J mol^{-1} K^{-1}, \log 0.0038=-2.42$)

Answer:

Explanation:

$\triangle G^{0}=-RT ln K$

$\triangle G^{0}=-2.303 RT \log K$

$\triangle G^{0}=-2.303 \times 8.314 mol^{-1} \times 298 \log 3.8 \times 10^{-3}$

$\triangle G^{0}=13.8 \times 10^{3} J mol ^{-1}$

$\triangle G^{0}=13.8 kJmol^{-1}$

A sample of argon of 1 atm pressure and 300 K expands reversibly and adiabatically from $1.25 dm^{3}$ to $2.5dm^{3}$ . Calculate  the approximate enthalpy (in J) change

(i) $C_{v}$ for argon is $12.48 JK^{-1}$

(ii) Assume argon to be an ideal gas

(iii) $\triangle T=111.5K$

Answer:

Explanation:

 We know that, $\triangle H=nC_{p} \triangle T$

 Number of moles of argon (n)= $\frac{pV}{RT}$

 given, p=1 atm, V=1.25 dm³, T=300 K

 $\therefore$   $n= \frac{1 \times 1.25}{0.082 \times 300}$=0.05 mol

 also, $C_{p}C_{v}=12.48 JK^{-1}$

 $C_{p}=C_{v}+R=12.48+8.314=20.794$

 On substituting  the above values in Eq.(i) ,

 we get

 $\triangle H=0.05 \times 20.794 \times 111.5=115.92$ ≈ 117 J

 If stoichiometric  quantites  of $KMnO_{4}$ and $K_{2}Cr_{2}O_{7}$ mixture is added for the oxidation  of $Fe^{2+}$ to $Fe^{3+}$ in acidic medium , then $Fe^{2+}$ will be oxidised

Answer:

Explanation:

 $KMnO_{4}+K_{2}Cr_{2}O_{7}$

 In the process $KMnO_{4}$ is strong oxidising agent in acidic medium . it oxidises $Fe^{2+} $ to $Fe^{3+}$ . $K_{2}Cr_{2}O_{7}$ is weak oxidising agent in acidic medium

 Find the odd-electron molecules from the following.

          $C_{2}H_{2}$,   $SCl_{2}$,   NO,   $N_{2}$ ,   $NO_{2}$

           as  (I)          ,    (ii)   ,    (iii)   ,    (iv)     ,     (v)     respectively

Answer:

Explanation:

 NO and $NO_{2}$  species are having odd electrons and disobey the octet rule.

 Formal charges and the molecule's resonance structure are indicated.

When uncertainty  in position  and momentum  are equal, then the uncertainty in velocity is 

Answer:

Explanation:

 $\triangle x= \triangle p$

 $\triangle p^{2}= \frac{h}{4 \pi}$     or   $\triangle p=\frac{1}{2}\sqrt{\frac{h}{\pi}}$

 $m\triangle v=\frac{1}{2}\sqrt{\frac{h}{\pi}}$  or,    $\triangle v=\frac{1}{2m}\sqrt{\frac{h}{\pi}}$

• Chemistry - General questions