QBManager

Chemical Equilibrium

229286 For the reaction ${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$ if we write $K_{p} = K_{c} (R T)^{x}$ , then $x$ becomes

1 1

2

- \frac{1}{2}

3

\frac{1}{2}

4 1

Explanation:

For the reaction -
${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$
$x =$ no. of mole of gaseous product - no. of mole of gaseous reactant
$x = 1 - (\frac{1}{2} + 1) = 1 - \frac{3}{2} = - \frac{1}{2}$

WB-JEE-2009

Chemical Equilibrium

229301 The equilibrium constant $K_{p}$ for the reaction, $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is :

1 more than one

2 less than one

3 equal to

K_{c}

4 zero

Explanation:

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 2 - 2 = 0$
$∴ K_{p} = K_{c} (RT)^{0} = K_{c}$

BCECE-2006

Chemical Equilibrium

229315 In the equilibrium reaction,
$A (g) + 2 B (g) ⇌ C (g) + D (g)$ , the equilibrium constant, $K_{c}$ is given by the expression

1

K_{c} = \frac{[C] [D]}{[A] [B]}

2

K_{c} = \frac{[C] [D]^{2}}{[A] [B]}

3

K_{c} = \frac{[A] [B]^{2}}{[C] [D]}

4

K_{c} = \frac{[C] [D]}{[A] [B]^{2}}

Explanation:

The equilibrium reaction is given below-
$A (g) + 2 B (g) ⇌ C (g) + D (g)$
The expression of equilibrium constant $(K_{c})$ for above reaction is-
$K_{c} = \frac{[C] [D]}{[A] [B]^{2}}$

AP-EAMCET-1994

Chemical Equilibrium

229286 For the reaction ${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$ if we write $K_{p} = K_{c} (R T)^{x}$ , then $x$ becomes

1 1

2

- \frac{1}{2}

3

\frac{1}{2}

4 1

Explanation:

For the reaction -
${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$
$x =$ no. of mole of gaseous product - no. of mole of gaseous reactant
$x = 1 - (\frac{1}{2} + 1) = 1 - \frac{3}{2} = - \frac{1}{2}$

WB-JEE-2009

Chemical Equilibrium

229301 The equilibrium constant $K_{p}$ for the reaction, $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is :

1 more than one

2 less than one

3 equal to

K_{c}

4 zero

Explanation:

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 2 - 2 = 0$
$∴ K_{p} = K_{c} (RT)^{0} = K_{c}$

BCECE-2006

Chemical Equilibrium

229315 In the equilibrium reaction,
$A (g) + 2 B (g) ⇌ C (g) + D (g)$ , the equilibrium constant, $K_{c}$ is given by the expression

1

K_{c} = \frac{[C] [D]}{[A] [B]}

2

K_{c} = \frac{[C] [D]^{2}}{[A] [B]}

3

K_{c} = \frac{[A] [B]^{2}}{[C] [D]}

4

K_{c} = \frac{[C] [D]}{[A] [B]^{2}}

Explanation:

The equilibrium reaction is given below-
$A (g) + 2 B (g) ⇌ C (g) + D (g)$
The expression of equilibrium constant $(K_{c})$ for above reaction is-
$K_{c} = \frac{[C] [D]}{[A] [B]^{2}}$

AP-EAMCET-1994

Chemical Equilibrium

229286 For the reaction ${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$ if we write $K_{p} = K_{c} (R T)^{x}$ , then $x$ becomes

1 1

2

- \frac{1}{2}

3

\frac{1}{2}

4 1

Explanation:

For the reaction -
${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3}$
$x =$ no. of mole of gaseous product - no. of mole of gaseous reactant
$x = 1 - (\frac{1}{2} + 1) = 1 - \frac{3}{2} = - \frac{1}{2}$

WB-JEE-2009

Chemical Equilibrium

229301 The equilibrium constant $K_{p}$ for the reaction, $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ is :

1 more than one

2 less than one

3 equal to

K_{c}

4 zero

Explanation:

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$
$K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 2 - 2 = 0$
$∴ K_{p} = K_{c} (RT)^{0} = K_{c}$

BCECE-2006

Chemical Equilibrium

229315 In the equilibrium reaction,
$A (g) + 2 B (g) ⇌ C (g) + D (g)$ , the equilibrium constant, $K_{c}$ is given by the expression

1

K_{c} = \frac{[C] [D]}{[A] [B]}

2

K_{c} = \frac{[C] [D]^{2}}{[A] [B]}

3

K_{c} = \frac{[A] [B]^{2}}{[C] [D]}

4

K_{c} = \frac{[C] [D]}{[A] [B]^{2}}

Explanation:

The equilibrium reaction is given below-
$A (g) + 2 B (g) ⇌ C (g) + D (g)$
The expression of equilibrium constant $(K_{c})$ for above reaction is-
$K_{c} = \frac{[C] [D]}{[A] [B]^{2}}$

AP-EAMCET-1994