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Chemistry Test - 16

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Weekly Quiz Competition

Question 1
4 / -1

Consider the following reactions:
\(\mathrm{NO}_{2} \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}+\mathrm{O}_{2}, \mathrm{K}_{1}\)
\(\mathrm{N}_{3} \mathrm{Q}_{4} \rightleftharpoons 2 \mathrm{N} \mathrm{O}_{2}, \mathrm{K}_{2}\)
Give the equilibrium constant for the formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) from \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\).

A
\(\frac{1}{k_{1}^{2}}+\frac{1}{k_{2}}\)

B
\(\frac{1}{K_{1}^{2} K_{2}}\)

C
\(\sqrt{\frac{1}{\mathrm{K}_{1} \mathrm{K}_{2}}}\)

D
\(\frac{K_{2}}{K_{1}}\)

Solution

\(N O_{2} \leftrightharpoons \frac{1}{2} N_{2}+O_{2}, K_{1}\)
\(N_{2}+2 O_{2} \leftrightharpoons 2 N O_{2}, K^{\prime}=\frac{1}{K_{1}^{2}} \rightarrow(i)\)
\(N_{2} O_{4}=2 N O_{2}, K_{2} \rightarrow(i i)\)
\((i)-(i i)\) willgiveus \(: N_{2}+2 O_{2} \leftrightharpoons N_{2} O_{4}, K_{3}=\frac{1}{K_{1}^{2} K_{2}}\)
Question 2
4 / -1

For the following reaction, 1 g mole of CaCO₃ is enclosed in a 5 L container.
CaCO₃ (s) CaO (s) + CO₂ (g)
Kp = 1.16 at 1073 K
The percent dissociation of CaCO₃ in the above reaction is

A
65%

B
100%

C
6.5%

D
zero

Solution

\(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)
\(K_{p}=P_{\mathrm{CO}_{2}} \quad\) (only gaseous molecule count) \(\begin{aligned} K_{p} &=K_{c}(R T)^{\Delta n} \\ \Rightarrow 1.16 &=\frac{x}{5}(0.0821 \times 1073)^{1} \\ x &=\frac{1.16 \times 5}{0.0821 \times 1073}=\frac{5.8}{88.1}=0.0658 \\ &=0.0658 \times 100=6.58 \% \end{aligned}\)
Question 3
4 / -1

For the reaction Cl₂ + 2I^-⇒I₂ + 2Cl^-, the initial concentration of I- was 0.20 mol L^-1 and the concentration after 20 min was 0.18 mol L^-1. The rate of formation of I2 in mol lit^-1 would be

A
1 * 10^-3

B
5 * 10^-4

C
1 * 10^-4

D
2 * 10^-3

Solution

Rate of formation of \(\mathrm{I}_{2}=\frac{d \mathrm{I}_{2}}{d t}=-\frac{1}{2} \frac{d \Gamma}{d t}\)
\[
\begin{array}{l}
=\frac{1}{2} \times \frac{0.20-0.18}{20}=\frac{1}{2} \times \frac{0.02}{20} \\
=5 \times 10^{-4} \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}
\end{array}
\]
Question 4
4 / -1

The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm⁻¹. The equivalent conductance of the solution is

A
3.14 mho⁻²cm²eq⁻¹

B
314.28 mho⁻¹cm²eq⁻¹

C
31.4 mho cm²eq⁻¹

D
314.28 mho cm²eq⁻¹

Solution

\(\lambda_{e q}=\frac{k \times 1000}{N} k=G \times \frac{l}{A}\)
\(k=\frac{1}{210} \times 0.66=0.00314286\)
\(\lambda_{e q}=\frac{k \times 1000}{n}=\frac{0.00314286 \times 1000}{0.01}=314.286 \mathrm{Scm}^{2} \mathrm{eq}^{-1}\)
Question 5
4 / -1

If 10²¹ molecules are removed from 200 mg of CO₂, then the number of moles of CO₂ left will be

A
2.88*10^-3

B
1.66*10^-3

C
4,54*10^-3

D
3.341*10^-3

Solution

\(200 \mathrm{mg} \mathrm{CO}_{2}=0.2 \mathrm{g}=\frac{0.2}{44} \mathrm{mol}\)
\(=0.00454 \mathrm{mol}=4.54 \times 10^{-3} \mathrm{mol}\)
\(10^{21}\) molecules of \(\mathrm{CO}_{2}=\frac{10^{21}}{6.02 \times 10^{23}}\)
\(=1.66 \times 10^{-3} \mathrm{mole}\)
\(\therefore\) No of moles left \(=(4.54-1.66) \times 10^{-3}\)
\(=2.88 \times 10^{-3}\)
Question 6
4 / -1

What is the relation between K_p and K_c in the following reaction?
N₂(g) + 3H₂(g) 2NH₃(g)

A
K_p = K_c

B
K_p = K_c - RT

C
K_p = K_c/(RT)²

D
K_p = K_c/(RT)^-2

Solution

The relation between \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{C}}\) is given as:
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}}\)
\(\mathrm{K}_{\mathrm{p}}=\) equilibrium constant in terms of partial pressure (its unit is 'atm'.) \(\mathrm{K}_{\mathrm{C}}=\) equilibrium constant in terms of concentration (molarity)
\(R=\) ideal gas constant \((0.082 \mathrm{atm} .4 / \mathrm{K} .\) mol \()\)
\(T=\) absolute temperature \((570 \mathrm{K})\) \(\Delta n=\Sigma\) n products \(-\Sigma\) n reactants \(\Delta n=(2)-(3+1)=-2\)
Thus, for the formation of \(\mathrm{NH}_{3}\), the correct relation between \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{C}}\) is:
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{-2} \mathrm{or}\)
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{O}} /(\mathrm{RT})^{2}\)
Question 7
4 / -1

How many coulombs of electricity is required for the complete reduction of MnO₄^- in 500 ml of 0.5 M solution to Mn₂⁺ ions?

A
96500 C

B
1.93 105 C

C
120625 C

D
9.65 * 10⁵ C

Solution

Moles of \(\mathrm{MnO}_{A}^{-}\) in \(500 \mathrm{ml}=500 \times 10^{-3} \times 0.5=0.25 \mathrm{mol}\)
\(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{M} \mathrm{n}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\)
\(1 \mathrm{mol}\)
\(5 \mathrm{mol}\)
1 mole of \(\mathrm{MnO}_{4}\) - require electricity \(=5 \times 96500 \mathrm{C}\) \(0.25 \mathrm{mol}\) of \(\mathrm{MnO}_{4}\) - require electricity \(=5 \times 96500 \times 0.25=120625 \mathrm{C}\)
Question 8
4 / -1

The pyrimidine bases present in DNA are-

A
Cytosine and adenine

B
Cytosine and guanine

C
Cytosine and thyamine

D
Cytosine and uracil

Solution

Pyrimidine is an aromatic heterocyclic organic compound similar to pyridine. In nucleic acids, three types of nucleobases are pyrimidine derivatives: cytosine (C), thymine (T), and uracil (U).
Question 9
4 / -1

Which of the following statements are true?
(A) Sucrose is a non-reducing sugar.
(B) Glucose is oxidised by bromine water.
(C) Glucose rotates plane polarised light in clockwise direction.
(D) Fructose is oxidised by bromine water.

A
(A), (B) and (C)

B
(A) and (B)

C
(B) and (C)

D
(A) and (D)

Solution

The statement (D) is incorrect. Fructose is a ketohexoses and is not oxidised by mild oxidising agents, like chlorine or bromine water.
All the other statements are correct.
(A) Sucrose has an anomeric carbon which is not free. The carbon that links glucose and fructose and fructose does not have free OH group to undergo reducing reaction and to open the ring, sucrose is non-reducing.
(B) With bromine water, glucose is oxidised to gluconic acid.
\(\mathrm{HOH}_{2} \mathrm{C}-(\mathrm{CHOH})_{4}-\mathrm{CHO} \longrightarrow \mathrm{HOCH}_{2}-(\mathrm{CHOH})_{4}-\mathrm{COOH}\)
(C) Glucose is dextrorotatory and rotates the plane polarised light in the clockwise direction.
Question 10
4 / -1

Standard enthalpies of formation of CO₂ (g), CO (g), N₂O (g) and NO₂(g) in kJ/mol are -393, -110, 81 and 34, respectively. What is the value of (in kJ) in the following reaction?
2NO₂ (g) + 3CO (g) \(\Rightarrow\) N₂O (g) + 3CO₂(g)

A
–846

B
–1460

C
–1460

D
836

Solution

Given the reaction: 2NO₂(g) + 3CO (g) \(\Rightarrow\) N₂O (g) + 3CO₂(g)
By using the formula:
D_H for the reaction = [Sum of the heat of formation for products] - [Sum of the heat of formation for reactants]
D_Hfor the reaction= [81-3×393]-[2×34-3×-110]=-846 KJ.
The value comes out to be –846 kJ.

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Chemistry Test - 16

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Chemistry Test - 16

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Question 1

\(\frac{1}{k_{1}^{2}}+\frac{1}{k_{2}}\)

\(\frac{1}{K_{1}^{2} K_{2}}\)

\(\sqrt{\frac{1}{\mathrm{K}_{1} \mathrm{K}_{2}}}\)

\(\frac{K_{2}}{K_{1}}\)

Solution

Question 2

65%

100%

6.5%

zero

Solution

Question 3

1 * 10-3

5 * 10-4

1 * 10-4

2 * 10-3

Solution

Question 4

3.14 mho−2cm2eq−1

314.28 mho−1cm2eq−1

31.4 mho cm2eq−1

314.28 mho cm2eq−1

Solution

Question 5

2.88*10-3

1.66*10-3

4,54*10-3

3.341*10-3

Solution

Question 6

Kp = Kc

Kp = Kc - RT

Kp = Kc/(RT)2

Kp = Kc/(RT)-2

Solution

Question 7

96500 C

1.93 105 C

120625 C

9.65 * 105 C

Solution

Question 8

Cytosine and adenine

Cytosine and guanine

Cytosine and thyamine

Cytosine and uracil

Solution

Question 9

(A), (B) and (C)

(A) and (B)

(B) and (C)

(A) and (D)

Solution

Question 10

–846

–1460

–1460

836

Solution

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1 * 10^-3

5 * 10^-4

1 * 10^-4

2 * 10^-3

3.14 mho⁻²cm²eq⁻¹

314.28 mho⁻¹cm²eq⁻¹

31.4 mho cm²eq⁻¹

314.28 mho cm²eq⁻¹

2.88*10^-3

1.66*10^-3

4,54*10^-3

3.341*10^-3

K_p = K_c

K_p = K_c - RT

K_p = K_c/(RT)²

K_p = K_c/(RT)^-2

9.65 * 10⁵ C