﻿ At 60° and 1 atm, N2O4 is 50% dissociated into NO2 then KP is from Chemistry NEET Year 2012 Free Solved Previous Year Papers

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# 1.At 60° and 1 atm, N2O4 is 50% dissociated into NO2 then KP is1.33 atm 2 atm 2.67 atm 3 atm

B.

2 atm

${\mathrm{N}}_{2}{\mathrm{O}}_{4}⇌2{\mathrm{NO}}_{2}$

Initially 1 0

At equili. 1-$\mathrm{\alpha }$ 2$\mathrm{\alpha }$

N2O4 is 50% dissociated, so$\mathrm{\alpha }$=$\frac{1}{2}$

${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{{\mathrm{NO}}_{2}}^{2}}{{\mathrm{p}}_{{\mathrm{N}}_{2}{\mathrm{O}}_{4}}}=\frac{\left(2×\frac{1}{2}{\right)}^{2}}{\left(1-\frac{1}{2}\right)}=2\mathrm{atm}$

2.

Which has the highest pH?

• CH3COOK

• Na2CO3

• NH4Cl

• NaNO3

B.

Na2CO3

NH4Cl solution is acidic, its pH < 7. NaNO3solution is neutral, its pH = 7. CH3COOK andNa2CO4solutions are basic their pH > 7. But Na2CO3solution is more basic, its pH > pH of CH3COOK solution.

3.

The dipole moment is minimum in

• NH3

• NF3

• SO2

• BF3

D.

BF3

BF3 has zero dipole moment

4.

Bond dissociation energy of CH4 is 360 kJ/mol and C2H6is 620 kJ/mol. Then bond dissociation energy of C-C bond is

• 170 kJ/mol

• 50 kJ/mol

• 80 kJ/mol

• 220 kJ/mol

C.

80 kJ/mol

Dissociation energy of methane = 360 kJ mol-1

Bond energy of C-H bond =$\frac{360}{4}=90\mathrm{kJ}$

Bond energy of ethane,

1(C-C) 6 (C-H) =620 kJ/mol

(C-C) + 6$×$90 = 620

(C-C) + 540 = 62

C-C = 80 kJ mo1-1
Bond dissociation of C-C bond= 80 kJ moJ-1.

5.

Arrange the following gases in order of their critical temperature.
NH3,H2O, CO2,O2

• NH> H2O > CO> O2

• O> CO> H2O > NH3

• H2O > NH> CO> O2

• CO>O>H2O > NH3

C.

H2O > NH> CO> O2

Greater are the intermolecular forces of attraction, higher is the critical temperature.

6.

At equilibrium which is correct?

• $\text{∆}$G = 0

• $\text{∆}$S = 0

• $\text{∆}$H = 0

•  $\text{∆}$${\text{G}}^{°}$= 0

A.

$\text{∆}$G = 0

In unfavorable reaction have Delta G values that are positive (also called endergonic reaction). When Delta G for reaction is zero, a reaction is said to be at equilibrium. Equilibrium does not mean equal concentrations. If the Delt G is zero, there is no net change in A and B, as the system is at equilibrium.

7.

Threshold frequency of a metal is 5$\text{×}$1013 s-1upon which 1 x 1014s-1frequency light is focused. Then the maximum kinetic energy of emitted electron is

• 3.3 $\text{×}$ 10-21

• 3.3 $\text{×}$ 10-20

• 6.6 $\text{×}$ 10-23

• 6.6 $\text{×}$ 10-20

B.

3.3 $\text{×}$ 10-20

Following the conservation of energy principle,

Kinetic energy$\left[\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^{2}\right]$= h($\text{v}$-${\mathrm{v}}_{\circ }$)

=(6.626$\text{×}$10-34) (1014s-1 - 5$\text{×}$1013 s-1)

=(6.626 x 10-34J s) (5 x 1013 s-1)

=3.313 x 10-20J.

8.

pKa increases in benzoic acid when substituent "x" is bonded at para-position, then "x" is

• -COOH

• -NO2

• -CN

• -OCH3

D.

-OCH3

Larger the value of pKa smaller will be its acidity. Out of the four groups, -COOH, -NO2and -CN are e-withdrawing which makes benzoic acid more acidic whereas -OCH3is e- donating which reduces the acidity (makes H less easily available). pKa value increases if OCH3 is present at para-position of benzoic acid.

9.

In Bohr's Orbit,$\frac{\mathrm{nh}}{2\mathrm{\pi }}$indicates

• Momentum

• Kinetic energy

• Potential energy

• Angular momentum

D.

Angular momentum

Angular momentum =$\mathrm{m\nu r}=\frac{\mathrm{nh}}{2\mathrm{\pi }}$

10.

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}$for following reaction will be${\mathrm{CO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}$.

• RT

• $\frac{1}{\mathrm{RT}}$

• $\frac{1}{\sqrt{\mathrm{RT}}}$

• $\frac{\mathrm{RT}}{2}$

C.

$\frac{1}{\sqrt{\mathrm{RT}}}$

${∆}_{{\mathrm{n}}_{\mathrm{g}}}={\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{r}}=1-\frac{3}{2}$

${∆}_{{\mathrm{n}}_{\mathrm{g}}}=\frac{-1}{2}$. Hence KP= Kc(RT)-1/2

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}=\frac{1}{\left(\mathrm{RT}{\right)}^{1/2}}=\frac{1}{\sqrt{\mathrm{RT}}}$