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Chemical Equilibrium

229060 The equilibrium constant for the reaction $H_{2} O (g) + CO (g)$ ) $H_{2} (g) + {CO}_{2} (g)$ is 64. The rate constant for the forward reaction is $160$ , the rate constant for the backward reaction is:

1 0.4

2 2.5

3 6.2

4

10.24 \times 10^{3}

Explanation:

Given values:- $K_{c} = 64, K_{f} = 160, K_{b} =$ ?
For the equilibrium reaction-
$H_{2} O + CO ⇌ H_{2} + {CO}_{2}$
At equilibrium-
$K_{c} = \frac{k_{f}}{k_{b}}$
Where- $K_{c} =$ Equilibrium constant
$k_{f} = Rate of forward reaction$
$k_{b} = Rate of backward reaction$
$k_{b} = \frac{160}{64}$
or $k_{b} = 2.5$

AP-EAMCET (Med.)-1999

Chemical Equilibrium

229061 The equilibrium constant $(K_{c})$ for the reaction $HA + B ⇌ {BH}^{+} + A^{-}$ is $100$ . If the rate constant for the forward reaction is $10^{5}$ , rate constant for the reverse reaction is

1

10^{- 3}

2

10^{- 5}

3

10^{7}

4

10^{3}

Explanation:

Given that, $K_{c} = 100, k_{f} = 10^{5}, k_{b} =$ ?
$HA + B ⇌ 栂 C (g) + D (g)$
At equilibrium-
$\begin{aligned} K_{c} & = \frac{k_{f}}{k_{b}} \\ or & k_{b} = \frac{k_{f}}{K_{c}} \end{aligned}$
Putting the value, we get-
$k_{b} = \frac{10^{5}}{100} = 10^{3}$

AP-EAMCET-1994

Chemical Equilibrium

229054 If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 K$ , at equilibrium $(K_{c} = 10^{- 6})$ , which one is correct?

1

[I_{2} (g)] > [I^{-} (g)]

2

[I_{2} (g)] < [I^{-} (g)]

3

[I_{2} (g)] = [I^{-} (g)]

4

[I_{2} (g)] = \frac{1}{2} [I^{-} (g)]

Explanation:

$\underset{1 - x}{I_{2}} ⇌ \underset{2 x}{2 I^{-}}$
$K_{c} = \frac{(2 x)^{2}}{(1 - x)} = 10^{- 6}$
It shows that $(1 - x) > 2 x \Rightarrow I_{2} > I^{-}$

BITSAT-2013

Chemical Equilibrium

229058 By applying law of mass action, the equilibrium constant $K$ for the reaction
$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$

1

K = \frac{[HA] [H_{2} O]}{[H_{3} O^{+}] [A^{-}]}

2

K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}

3

K = \frac{[H_{3} O^{+}] [H_{2} O]}{[A^{-}] [HA]}

4

K = \frac{[HA] [A^{-}]}{[H_{2} O] [H_{3} O^{+}]}

Explanation:

The reaction is :
$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$
The equilibrium constant $K = \frac{[Product]}{[Reactant]}$
So, $K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}$

J and K CET-(2007)

Chemical Equilibrium

229060 The equilibrium constant for the reaction $H_{2} O (g) + CO (g)$ ) $H_{2} (g) + {CO}_{2} (g)$ is 64. The rate constant for the forward reaction is $160$ , the rate constant for the backward reaction is:

1 0.4

2 2.5

3 6.2

4

10.24 \times 10^{3}

Explanation:

Given values:- $K_{c} = 64, K_{f} = 160, K_{b} =$ ?
For the equilibrium reaction-
$H_{2} O + CO ⇌ H_{2} + {CO}_{2}$
At equilibrium-
$K_{c} = \frac{k_{f}}{k_{b}}$
Where- $K_{c} =$ Equilibrium constant
$k_{f} = Rate of forward reaction$
$k_{b} = Rate of backward reaction$
$k_{b} = \frac{160}{64}$
or $k_{b} = 2.5$

AP-EAMCET (Med.)-1999

Chemical Equilibrium

229061 The equilibrium constant $(K_{c})$ for the reaction $HA + B ⇌ {BH}^{+} + A^{-}$ is $100$ . If the rate constant for the forward reaction is $10^{5}$ , rate constant for the reverse reaction is

1

10^{- 3}

2

10^{- 5}

3

10^{7}

4

10^{3}

Explanation:

Given that, $K_{c} = 100, k_{f} = 10^{5}, k_{b} =$ ?
$HA + B ⇌ 栂 C (g) + D (g)$
At equilibrium-
$\begin{aligned} K_{c} & = \frac{k_{f}}{k_{b}} \\ or & k_{b} = \frac{k_{f}}{K_{c}} \end{aligned}$
Putting the value, we get-
$k_{b} = \frac{10^{5}}{100} = 10^{3}$

AP-EAMCET-1994

Chemical Equilibrium

229054 If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 K$ , at equilibrium $(K_{c} = 10^{- 6})$ , which one is correct?

1

[I_{2} (g)] > [I^{-} (g)]

2

[I_{2} (g)] < [I^{-} (g)]

3

[I_{2} (g)] = [I^{-} (g)]

4

[I_{2} (g)] = \frac{1}{2} [I^{-} (g)]

Explanation:

$\underset{1 - x}{I_{2}} ⇌ \underset{2 x}{2 I^{-}}$
$K_{c} = \frac{(2 x)^{2}}{(1 - x)} = 10^{- 6}$
It shows that $(1 - x) > 2 x \Rightarrow I_{2} > I^{-}$

BITSAT-2013

Chemical Equilibrium

229058 By applying law of mass action, the equilibrium constant $K$ for the reaction
$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$

1

K = \frac{[HA] [H_{2} O]}{[H_{3} O^{+}] [A^{-}]}

2

K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}

3

K = \frac{[H_{3} O^{+}] [H_{2} O]}{[A^{-}] [HA]}

4

K = \frac{[HA] [A^{-}]}{[H_{2} O] [H_{3} O^{+}]}

Explanation:

The reaction is :
$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$
The equilibrium constant $K = \frac{[Product]}{[Reactant]}$
So, $K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}$

J and K CET-(2007)

Chemical Equilibrium

229060 The equilibrium constant for the reaction $H_{2} O (g) + CO (g)$ ) $H_{2} (g) + {CO}_{2} (g)$ is 64. The rate constant for the forward reaction is $160$ , the rate constant for the backward reaction is:

1 0.4

2 2.5

3 6.2

4

10.24 \times 10^{3}

Explanation:

Given values:- $K_{c} = 64, K_{f} = 160, K_{b} =$ ?
For the equilibrium reaction-
$H_{2} O + CO ⇌ H_{2} + {CO}_{2}$
At equilibrium-
$K_{c} = \frac{k_{f}}{k_{b}}$
Where- $K_{c} =$ Equilibrium constant
$k_{f} = Rate of forward reaction$
$k_{b} = Rate of backward reaction$
$k_{b} = \frac{160}{64}$
or $k_{b} = 2.5$

AP-EAMCET (Med.)-1999

Chemical Equilibrium

229061 The equilibrium constant $(K_{c})$ for the reaction $HA + B ⇌ {BH}^{+} + A^{-}$ is $100$ . If the rate constant for the forward reaction is $10^{5}$ , rate constant for the reverse reaction is

1

10^{- 3}

2

10^{- 5}

3

10^{7}

4

10^{3}

Explanation:

Given that, $K_{c} = 100, k_{f} = 10^{5}, k_{b} =$ ?
$HA + B ⇌ 栂 C (g) + D (g)$
At equilibrium-
$\begin{aligned} K_{c} & = \frac{k_{f}}{k_{b}} \\ or & k_{b} = \frac{k_{f}}{K_{c}} \end{aligned}$
Putting the value, we get-
$k_{b} = \frac{10^{5}}{100} = 10^{3}$

AP-EAMCET-1994

Chemical Equilibrium

229054 If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 K$ , at equilibrium $(K_{c} = 10^{- 6})$ , which one is correct?

1

[I_{2} (g)] > [I^{-} (g)]

2

[I_{2} (g)] < [I^{-} (g)]

3

[I_{2} (g)] = [I^{-} (g)]

4

[I_{2} (g)] = \frac{1}{2} [I^{-} (g)]

Explanation:

$\underset{1 - x}{I_{2}} ⇌ \underset{2 x}{2 I^{-}}$
$K_{c} = \frac{(2 x)^{2}}{(1 - x)} = 10^{- 6}$
It shows that $(1 - x) > 2 x \Rightarrow I_{2} > I^{-}$

BITSAT-2013

Chemical Equilibrium

229058 By applying law of mass action, the equilibrium constant $K$ for the reaction
$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$

1

K = \frac{[HA] [H_{2} O]}{[H_{3} O^{+}] [A^{-}]}

2

K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}

3

K = \frac{[H_{3} O^{+}] [H_{2} O]}{[A^{-}] [HA]}

4

K = \frac{[HA] [A^{-}]}{[H_{2} O] [H_{3} O^{+}]}

Explanation:

The reaction is :
$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$
The equilibrium constant $K = \frac{[Product]}{[Reactant]}$
So, $K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}$

J and K CET-(2007)

Chemical Equilibrium

229060 The equilibrium constant for the reaction $H_{2} O (g) + CO (g)$ ) $H_{2} (g) + {CO}_{2} (g)$ is 64. The rate constant for the forward reaction is $160$ , the rate constant for the backward reaction is:

1 0.4

2 2.5

3 6.2

4

10.24 \times 10^{3}

Explanation:

Given values:- $K_{c} = 64, K_{f} = 160, K_{b} =$ ?
For the equilibrium reaction-
$H_{2} O + CO ⇌ H_{2} + {CO}_{2}$
At equilibrium-
$K_{c} = \frac{k_{f}}{k_{b}}$
Where- $K_{c} =$ Equilibrium constant
$k_{f} = Rate of forward reaction$
$k_{b} = Rate of backward reaction$
$k_{b} = \frac{160}{64}$
or $k_{b} = 2.5$

AP-EAMCET (Med.)-1999

Chemical Equilibrium

229061 The equilibrium constant $(K_{c})$ for the reaction $HA + B ⇌ {BH}^{+} + A^{-}$ is $100$ . If the rate constant for the forward reaction is $10^{5}$ , rate constant for the reverse reaction is

1

10^{- 3}

2

10^{- 5}

3

10^{7}

4

10^{3}

Explanation:

Given that, $K_{c} = 100, k_{f} = 10^{5}, k_{b} =$ ?
$HA + B ⇌ 栂 C (g) + D (g)$
At equilibrium-
$\begin{aligned} K_{c} & = \frac{k_{f}}{k_{b}} \\ or & k_{b} = \frac{k_{f}}{K_{c}} \end{aligned}$
Putting the value, we get-
$k_{b} = \frac{10^{5}}{100} = 10^{3}$

AP-EAMCET-1994

Chemical Equilibrium

229054 If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 K$ , at equilibrium $(K_{c} = 10^{- 6})$ , which one is correct?

1

[I_{2} (g)] > [I^{-} (g)]

2

[I_{2} (g)] < [I^{-} (g)]

3

[I_{2} (g)] = [I^{-} (g)]

4

[I_{2} (g)] = \frac{1}{2} [I^{-} (g)]

Explanation:

$\underset{1 - x}{I_{2}} ⇌ \underset{2 x}{2 I^{-}}$
$K_{c} = \frac{(2 x)^{2}}{(1 - x)} = 10^{- 6}$
It shows that $(1 - x) > 2 x \Rightarrow I_{2} > I^{-}$

BITSAT-2013

Chemical Equilibrium

229058 By applying law of mass action, the equilibrium constant $K$ for the reaction
$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$

1

K = \frac{[HA] [H_{2} O]}{[H_{3} O^{+}] [A^{-}]}

2

K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}

3

K = \frac{[H_{3} O^{+}] [H_{2} O]}{[A^{-}] [HA]}

4

K = \frac{[HA] [A^{-}]}{[H_{2} O] [H_{3} O^{+}]}

Explanation:

The reaction is :
$HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}$
The equilibrium constant $K = \frac{[Product]}{[Reactant]}$
So, $K = \frac{[H_{3} O^{+}] [A^{-}]}{[HA] [H_{2} O]}$

J and K CET-(2007)