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Chemical Equilibrium

NEET Chemistry · 93 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
In which one of the following equilibria, Kp ≠ \ne KC ?
A 2NO(g) ⇋ N2(g) + O2(g)
B 2C(s) + O2(g) ⇋ 2CO(g)
C 2HI(g) ⇋ H2(g) + I2(g)
D NO2(g) + SO2(g) ⇋ NO(g) + SO3(g)
Correct Answer
Option B
Solution

We know, Kp = KC(RT)

Δ\Delta

ng Kp = KC when

Δ\Delta

ng = 0 Kp ≠\ne KC when

Δ\Delta

ng ≠\ne 0 and T ≠\ne 12 K In this reaction, 2C(s) + O2(g) ⇋ 2CO(g)

Δ\Delta

ng = 1, so Kp ≠\ne KC

Q42
The equilibrium constant for the reaction SO3( g)⇌SO2( g)+12O2( g)\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\dfrac{1}{2} \mathrm{O}_2(\mathrm{~g}) is Kc=4.9×10−2\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}. The value of Kc\mathrm{K}_{\mathrm{c}} for the reaction given below is 2SO2( g)+O2( g)⇌2SO3( g)2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) is :
A 49
B 416
C 41.6
D 4.9
Correct Answer
Option B
Solution

The reaction

2SO2+O2⇌2SO32 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3

can be formed from the given reaction by reverting it and multiplying coefficients by 2. Thus,

Kc′=Kc−2=1 Kc2=(14.9×10−2)2=416\begin{aligned} \mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =416 \end{aligned}
Q43
For the given hypothetical reactions, the equilibrium constants are as follows : X⇌Y;K1=1.0Y⇌Z;K2=2.0Z⇌W;K3=4.0\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0 \end{aligned} The equilibrium constant for the reaction X⇌W\mathrm{X} \rightleftharpoons \mathrm{W} is
A 12.0
B 8.0
C 6.0
D 7.0
Correct Answer
Option B
Solution

To find the equilibrium constant for the overall reaction

X⇌W\mathrm{X} \rightleftharpoons \mathrm{W}

, we need to combine the equilibrium constants for the individual reactions given.

Let's analyze this step-by-step: The given reactions and their equilibrium constants are:

X⇌YK1=1.0Y⇌ZK2=2.0Z⇌WK3=4.0\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} \quad K_1 = 1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} \quad K_2 = 2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} \quad K_3 = 4.0 \end{aligned}

The equilibrium constant for the overall reaction

X⇌W\mathrm{X} \rightleftharpoons \mathrm{W}

is the product of the equilibrium constants for the individual steps.

This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.

So, we have:

Koverall=K1⋅K2⋅K3K_{\text{overall}} = K_1 \cdot K_2 \cdot K_3

Now, substituting the given values:

Koverall=1.0⋅2.0⋅4.0=8.0K_{\text{overall}} = 1.0 \cdot 2.0 \cdot 4.0 = 8.0

Therefore, the equilibrium constant for the reaction

X⇌W\mathrm{X} \rightleftharpoons \mathrm{W}

is 8.0, which corresponds to Option B.

Q44
The ratio KPKC\dfrac{K_P}{K_C} for the reaction : CO(g)+12O2( g)⇌CO2( g)\mathrm{CO}_{(\mathrm{g})}+\dfrac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})} is :
A 1
B (RT)1/2 (\mathrm{RT})^{1 / 2}
C RT
D 1RT\mathrm{ \dfrac{1}{\sqrt{R T}}}
Correct Answer
Option D
Solution

To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure,

KPK_P

, and the equilibrium constant in terms of concentration,

KCK_C

. The relationship between these two constants for a general reaction is given by:

KP=KC(RT)ΔnK_P = K_C (RT)^{\Delta n}

where:

Δn\Delta n

is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),

RR

is the ideal gas constant, and

TT

is the temperature in Kelvin. For the given reaction:

CO(g)+12O2( g)⇌CO2( g)\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}

we need to calculate

Δn\Delta n

. On the reactants side, we have: 1 mole of

CO\mathrm{CO}

(g) 0.5 moles of

O2\mathrm{O}_2

(g) So, the total number of moles of reactants is:

1+12=32=1.51 + \frac{1}{2} = \frac{3}{2} = 1.5

On the products side, we have: 1 mole of

CO2\mathrm{CO}_2

(g) The total number of moles of products is:

11

Therefore,

Δn\Delta n

is:

Δn=moles of products−moles of reactants=1−1.5=−0.5\Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 1.5 = -0.5

Now, we can use the relationship between

KPK_P

and

KCK_C

:

KP=KC(RT)ΔnK_P = K_C (RT)^{\Delta n}

Substituting

Δn=−0.5\Delta n = -0.5

, we get:

KP=KC(RT)−0.5K_P = K_C (RT)^{-0.5}

or equivalently:

KPKC=(RT)−0.5\frac{K_P}{K_C} = (RT)^{-0.5}

Therefore, the correct option is: Option D:

1RT\frac{1}{\sqrt{R T}}
Q45
An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules :
A leaves the solution increases
B leaves the vapour increases
C leaves the vapour decreases
D leaves the solution decreases
Correct Answer
Option D
Solution

With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure.

Q46
A vessel at 1000 K contains CO2\mathrm{CO}_2 with a pressure of 0.5 atm . Some of CO2\mathrm{CO}_2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then Kp is :
A 0.18 atm
B 0.3 atm
C 3 atm
D 1.8 atm
Correct Answer
Option D
Solution

The reaction is:

CO2(g)+C(s)⇌2CO(g)\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2\mathrm{CO}(g)

Initially, the pressure of CO2\mathrm{CO}_2 is 0.5 atm, and the pressure of CO\mathrm{CO} is 0 atm.

Let 'x' be the change in pressure of CO2\mathrm{CO}_2 at equilibrium.

Since the stoichiometric coefficient of CO is twice that of CO2\mathrm{CO}_2, the pressure of CO formed at equilibrium is 2x.

At equilibrium: Pressure of CO2=0.5−x\mathrm{CO}_2 = 0.5 - x Pressure of CO=2x\mathrm{CO} = 2x The total pressure at equilibrium is given as 0.8 atm.

Therefore:

(0.5−x)+2x=0.8(0.5 - x) + 2x = 0.8
0.5+x=0.80.5 + x = 0.8
x=0.8−0.5=0.3x = 0.8 - 0.5 = 0.3

So, at equilibrium: Pressure of CO2=0.5−0.3=0.2 atm\mathrm{CO}_2 = 0.5 - 0.3 = 0.2 \text{ atm} Pressure of CO=2(0.3)=0.6 atm\mathrm{CO} = 2(0.3) = 0.6 \text{ atm} The equilibrium constant KpK_p is given by:

Kp=PCO2PCO2K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}
Kp=(0.6)20.2=0.360.2=1.8 atmK_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \text{ atm}

Therefore, the value of KpK_p is 1.8 atm.

Q47
Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(g)⇌B(g)+C(g)\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) If pp being the total pressure, KpK_p is the pressure equilibrium constant and α\alpha is the degree of dissociation, then which of the following is true at equilibrium?
A If KpK_p value is extremely high compared to p,αp, \alpha becomes much less than unity
B When p increases α\alpha increases
C If p value is extremely high compared to Kp,α≈1\mathrm{K}_{\mathrm{p}}, \alpha \approx 1
D When pp increases α\alpha decreases
Correct Answer
Option D
Solution

a moles of A(g)A(g) taken initially and at time Now moles fraction of A(g),B(g)\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g}) and C(g)\mathrm{C}(\mathrm{g}) are

XA=a−aαa+aα=1−α1+αXB=aαa+aα=α1+αXC=aαa+aα=α1+α\begin{aligned} & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{a}-\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{1-\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{B}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \end{aligned}

Now if P is total pressure then partial pressure of A(g),B(g)\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g}) and C(g)\mathrm{C}(\mathrm{g}) are

PA=(1−α1+α)PPB=(α1+α)PPC=(α1+α)P\begin{aligned} & \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \end{aligned}
KP=(α1+α)P(α1+α)P(1−α1+α)P KP=α2P1−α2\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}} \\ & \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2} \end{aligned}

As KPK_P is only function of temperature. So as P↑α↓\mathrm{P} \uparrow \quad \alpha \downarrow

Q48
For the reaction, CO(g) + Cl2(g) ⇋\leftrightharpoons COCl2(g) the KpKc{{{K_p}} \over {{K_c}}} is equal to :
A RT\sqrt {RT}
B RT
C 1/RT
D 1.0
Correct Answer
Option C
Solution
Kp=Kc(RT)Δn;{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};

Here

Δn=1−2=−1\Delta n = 1 - 2 = - 1

∴\therefore

kpKc=1RT{{{k_p}} \over {{K_c}}} = {1 \over {RT}}
Q49
At −20∘C-20^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} pressure, a cylinder is filled with equal number of H2,I2\mathrm{H}_2, \mathrm{I}_2 and HI\mathrm{HI} molecules for the reaction H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), the Kp\mathrm{K}_{\mathrm{p}} for the process is x×10−1x \times 10^{-1}. x=\mathrm{x}= __________. [Given : R=0.082 L atm K−1 mol−1\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}]
A 2
B 1
C 10
D 0.01
Correct Answer
Option C
Solution

At

−20∘C-20^{\circ} \mathrm{C}

and a pressure of

1 atm1 \mathrm{~atm}

, a cylinder contains equal quantities of

H2,I2,\mathrm{H}_2, \mathrm{I}_2,

and

HI\mathrm{HI}

molecules. The equilibrium constant for the reaction

H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})

denoted as

Kp\mathrm{K}_{\mathrm{p}}

, is given by the expression

x×10−1x \times 10^{-1}

. To solve for

xx

, use the provided equilibrium expression:

KP=KC=[HI]2[H2][I2]=1=10×10−1\mathrm{K_{P} = \mathrm{K_C}} =\frac{\left[\mathrm{HI}\right]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=1 = 10 \times 10^{-1}

Therefore,

x=10x = 10

.

Q50
The equilibrium constant for the reaction SO3 (g) ⇋\leftrightharpoons SO2 (g) + 121 \over 2 O2 (g) is Kc = 4.9 ×\times 10–2. The value of Kc for the reaction 2SO2 (g) + O2 (g) ⇋\leftrightharpoons 2SO3 (g) will be :
A 416
B 9.8 ×\times 10-2
C 4.9 ×\times 10-2
D 2.40 ×\times 10-3
Correct Answer
Option A
Solution
SO3(g) ⇌ SO2(g)  +  12O2(g)S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)
Kc=[SO2][O2]1/2[SO3]{K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{{1}/{2}}}}} \over {\left[ {S{O_3}} \right]}}
=4.9×10−2;= 4.9 \times {10^{ - 2}};

On taking the square of the above reaction

[SO2]2[O2][SO3]2{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}
=24.01×10−4= 24.01 \times {10^{ - 4}}

now

KC′K{'_C}

for

  2SO2(g)  +  O2(g) ⇌ 2SO3\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}
=[SO3]2[SO2]2[O2]= {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}
=124.01×10−4= {1 \over {24.01 \times {{10}^{ - 4}}}}
=416= 416
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