QBManager

CHXI07:EQUILIBRIUM

314465 28g of $N_{2}$ and $6 g$ of $H_{2}$ were mixed. At equilibrium $17 g$ ${NH}_{3}$ was produced. The weights of $N_{2}$ and $H_{2}$ at equilibrium are respectively

1

11 g, 0 g

2

1 g, 3 g

3

14 g, 3 g

4 11g,3g

Explanation:

$\begin{matrix} N_{2} & + & 3 H_{2} & ⇌ & 2 N H_{3} \\ Initially & 28 g & 6 g & 0 \\ 1 mole & 3 mole & 2 mole \\ At equil & (1 - 0.5) & (3 - 1.5) & 17 g N H_{3} \end{matrix}$
Concentration of 1 mole of $N_{2}$ combines with 3 moles of $H_{2}$ to produce 2 moles of ${NH}_{3}$ but, since at equilibrium 1 mole of ${NH}_{3}$ is produced. So, 0.5 moles, i.e. $14 g$ of $N_{2}$ are left at equilibrium, Similarly, 3 moles of $H_{2} \equiv 2$ moles of ${NH}_{3}$ .
For 1 mole of ${NH}_{3} \equiv 1.5$ mole of $H_{2}$
Hence, 1.5 mole i.e. $3 g$ of $H_{2}$ are left at equilibrium.
So, $14 g$ of $N_{2}$ and 3g of $H_{2}$ are left at equilibrium.

CHXI07:EQUILIBRIUM

314466 4 moles each of ${SO}_{2}$ and $O_{2}$ gases are allowed to react to form ${SO}_{3}$ in a closed vessel. At equilibrium $25 %$ of $O_{2}$ is used up. The total number of moles of all the gases at equilibrium is

1 6.5

2 7.0

3 8.0

4 2.0

Explanation:

supporting img
Total number of moles $= 2 + 3 + 2 = 7$ moles.

KCET - 2006

CHXI07:EQUILIBRIUM

314489 Assertion :
The value of $K$ gives us a relative idea about the extent to which a reaction proceeds.
Reason :
The value of $K$ is independent of the stoichiometry of reactants and products at the point of equilibrium.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXI07:EQUILIBRIUM

314468 Three moles of ${PCl}_{5}$ , three moles of ${PCl}_{3}$ and two moles of ${Cl}_{2}$ are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of ${PCl}_{5}$ , the number of moles of ${PCl}_{3}$ present in it is

1 6

2 4.5

3 5

4 3

Explanation:

$\begin{matrix} PC l_{5(g)} ⇌ & PC l_{3(g)} & + & C l_{2(g)} \\ Initially: & 3 & 3 & 2 \\ At eqm: & (3 - x) & (3 + x) & (2 + x) \end{matrix}$
Given that, $(3 - x) = 1.5 \Rightarrow x = 1.5$
Therefore no. of moles of ${PCl}_{3 (g)}$ present is
$= (3 + x) = 3 + 1.5 = 4.5$

KCET - 2008

CHXI07:EQUILIBRIUM

314465 28g of $N_{2}$ and $6 g$ of $H_{2}$ were mixed. At equilibrium $17 g$ ${NH}_{3}$ was produced. The weights of $N_{2}$ and $H_{2}$ at equilibrium are respectively

1

11 g, 0 g

2

1 g, 3 g

3

14 g, 3 g

4 11g,3g

Explanation:

$\begin{matrix} N_{2} & + & 3 H_{2} & ⇌ & 2 N H_{3} \\ Initially & 28 g & 6 g & 0 \\ 1 mole & 3 mole & 2 mole \\ At equil & (1 - 0.5) & (3 - 1.5) & 17 g N H_{3} \end{matrix}$
Concentration of 1 mole of $N_{2}$ combines with 3 moles of $H_{2}$ to produce 2 moles of ${NH}_{3}$ but, since at equilibrium 1 mole of ${NH}_{3}$ is produced. So, 0.5 moles, i.e. $14 g$ of $N_{2}$ are left at equilibrium, Similarly, 3 moles of $H_{2} \equiv 2$ moles of ${NH}_{3}$ .
For 1 mole of ${NH}_{3} \equiv 1.5$ mole of $H_{2}$
Hence, 1.5 mole i.e. $3 g$ of $H_{2}$ are left at equilibrium.
So, $14 g$ of $N_{2}$ and 3g of $H_{2}$ are left at equilibrium.

CHXI07:EQUILIBRIUM

314466 4 moles each of ${SO}_{2}$ and $O_{2}$ gases are allowed to react to form ${SO}_{3}$ in a closed vessel. At equilibrium $25 %$ of $O_{2}$ is used up. The total number of moles of all the gases at equilibrium is

1 6.5

2 7.0

3 8.0

4 2.0

Explanation:

supporting img
Total number of moles $= 2 + 3 + 2 = 7$ moles.

KCET - 2006

CHXI07:EQUILIBRIUM

314489 Assertion :
The value of $K$ gives us a relative idea about the extent to which a reaction proceeds.
Reason :
The value of $K$ is independent of the stoichiometry of reactants and products at the point of equilibrium.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXI07:EQUILIBRIUM

314468 Three moles of ${PCl}_{5}$ , three moles of ${PCl}_{3}$ and two moles of ${Cl}_{2}$ are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of ${PCl}_{5}$ , the number of moles of ${PCl}_{3}$ present in it is

1 6

2 4.5

3 5

4 3

Explanation:

$\begin{matrix} PC l_{5(g)} ⇌ & PC l_{3(g)} & + & C l_{2(g)} \\ Initially: & 3 & 3 & 2 \\ At eqm: & (3 - x) & (3 + x) & (2 + x) \end{matrix}$
Given that, $(3 - x) = 1.5 \Rightarrow x = 1.5$
Therefore no. of moles of ${PCl}_{3 (g)}$ present is
$= (3 + x) = 3 + 1.5 = 4.5$

KCET - 2008

CHXI07:EQUILIBRIUM

314465 28g of $N_{2}$ and $6 g$ of $H_{2}$ were mixed. At equilibrium $17 g$ ${NH}_{3}$ was produced. The weights of $N_{2}$ and $H_{2}$ at equilibrium are respectively

1

11 g, 0 g

2

1 g, 3 g

3

14 g, 3 g

4 11g,3g

Explanation:

$\begin{matrix} N_{2} & + & 3 H_{2} & ⇌ & 2 N H_{3} \\ Initially & 28 g & 6 g & 0 \\ 1 mole & 3 mole & 2 mole \\ At equil & (1 - 0.5) & (3 - 1.5) & 17 g N H_{3} \end{matrix}$
Concentration of 1 mole of $N_{2}$ combines with 3 moles of $H_{2}$ to produce 2 moles of ${NH}_{3}$ but, since at equilibrium 1 mole of ${NH}_{3}$ is produced. So, 0.5 moles, i.e. $14 g$ of $N_{2}$ are left at equilibrium, Similarly, 3 moles of $H_{2} \equiv 2$ moles of ${NH}_{3}$ .
For 1 mole of ${NH}_{3} \equiv 1.5$ mole of $H_{2}$
Hence, 1.5 mole i.e. $3 g$ of $H_{2}$ are left at equilibrium.
So, $14 g$ of $N_{2}$ and 3g of $H_{2}$ are left at equilibrium.

CHXI07:EQUILIBRIUM

314466 4 moles each of ${SO}_{2}$ and $O_{2}$ gases are allowed to react to form ${SO}_{3}$ in a closed vessel. At equilibrium $25 %$ of $O_{2}$ is used up. The total number of moles of all the gases at equilibrium is

1 6.5

2 7.0

3 8.0

4 2.0

Explanation:

supporting img
Total number of moles $= 2 + 3 + 2 = 7$ moles.

KCET - 2006

CHXI07:EQUILIBRIUM

314489 Assertion :
The value of $K$ gives us a relative idea about the extent to which a reaction proceeds.
Reason :
The value of $K$ is independent of the stoichiometry of reactants and products at the point of equilibrium.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXI07:EQUILIBRIUM

314468 Three moles of ${PCl}_{5}$ , three moles of ${PCl}_{3}$ and two moles of ${Cl}_{2}$ are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of ${PCl}_{5}$ , the number of moles of ${PCl}_{3}$ present in it is

1 6

2 4.5

3 5

4 3

Explanation:

$\begin{matrix} PC l_{5(g)} ⇌ & PC l_{3(g)} & + & C l_{2(g)} \\ Initially: & 3 & 3 & 2 \\ At eqm: & (3 - x) & (3 + x) & (2 + x) \end{matrix}$
Given that, $(3 - x) = 1.5 \Rightarrow x = 1.5$
Therefore no. of moles of ${PCl}_{3 (g)}$ present is
$= (3 + x) = 3 + 1.5 = 4.5$

KCET - 2008

CHXI07:EQUILIBRIUM

314465 28g of $N_{2}$ and $6 g$ of $H_{2}$ were mixed. At equilibrium $17 g$ ${NH}_{3}$ was produced. The weights of $N_{2}$ and $H_{2}$ at equilibrium are respectively

1

11 g, 0 g

2

1 g, 3 g

3

14 g, 3 g

4 11g,3g

Explanation:

$\begin{matrix} N_{2} & + & 3 H_{2} & ⇌ & 2 N H_{3} \\ Initially & 28 g & 6 g & 0 \\ 1 mole & 3 mole & 2 mole \\ At equil & (1 - 0.5) & (3 - 1.5) & 17 g N H_{3} \end{matrix}$
Concentration of 1 mole of $N_{2}$ combines with 3 moles of $H_{2}$ to produce 2 moles of ${NH}_{3}$ but, since at equilibrium 1 mole of ${NH}_{3}$ is produced. So, 0.5 moles, i.e. $14 g$ of $N_{2}$ are left at equilibrium, Similarly, 3 moles of $H_{2} \equiv 2$ moles of ${NH}_{3}$ .
For 1 mole of ${NH}_{3} \equiv 1.5$ mole of $H_{2}$
Hence, 1.5 mole i.e. $3 g$ of $H_{2}$ are left at equilibrium.
So, $14 g$ of $N_{2}$ and 3g of $H_{2}$ are left at equilibrium.

CHXI07:EQUILIBRIUM

314466 4 moles each of ${SO}_{2}$ and $O_{2}$ gases are allowed to react to form ${SO}_{3}$ in a closed vessel. At equilibrium $25 %$ of $O_{2}$ is used up. The total number of moles of all the gases at equilibrium is

1 6.5

2 7.0

3 8.0

4 2.0

Explanation:

supporting img
Total number of moles $= 2 + 3 + 2 = 7$ moles.

KCET - 2006

CHXI07:EQUILIBRIUM

314489 Assertion :
The value of $K$ gives us a relative idea about the extent to which a reaction proceeds.
Reason :
The value of $K$ is independent of the stoichiometry of reactants and products at the point of equilibrium.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXI07:EQUILIBRIUM

314468 Three moles of ${PCl}_{5}$ , three moles of ${PCl}_{3}$ and two moles of ${Cl}_{2}$ are taken in a closed vessel. If at equilibrium the vessel has 1.5 moles of ${PCl}_{5}$ , the number of moles of ${PCl}_{3}$ present in it is

1 6

2 4.5

3 5

4 3

Explanation:

$\begin{matrix} PC l_{5(g)} ⇌ & PC l_{3(g)} & + & C l_{2(g)} \\ Initially: & 3 & 3 & 2 \\ At eqm: & (3 - x) & (3 + x) & (2 + x) \end{matrix}$
Given that, $(3 - x) = 1.5 \Rightarrow x = 1.5$
Therefore no. of moles of ${PCl}_{3 (g)}$ present is
$= (3 + x) = 3 + 1.5 = 4.5$

KCET - 2008

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