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Byju's Answer

Standard XII

Chemistry

Degree of Dissociation

2 mol of N2...

Question

$2$ mol of $N_{2}$ is mixed with $6$ mol of $H_{2}$ in a closed vessel of one litre capacity. If $50$ % $N_{2}$ is converted into $N H_{3}$ at equilibrium, the value of $K_{C}$ for the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is:

4 / 27

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Solution

The correct option is A $4 / 27$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
at $t = 0$ 2mol 6mol $-$
at equilibrium 50% $N_{2}$ is converted into ${N H}_{3}$ .
at equim, (2-1) 3 2
$K_{C} = \frac{{[{N H}_{3}]}^{2}}{{[H_{2}]}^{3} [N_{2}]} = \frac{{(2)}^{2}}{{(3)}^{3} \times 1} = \frac{4}{27}$

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Similar questions

2

mol of

N_{2}

is mixed with

6

mol of

H_{2}

in a closed vessel of one litre capacity. If

50 %

N_{2}

is converted into

N H_{3}

at equilibrium, the value of

K_{c}

for the reaction.

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

Q. 2 mol of

N_{2}

is mixed with 6 mol of

H_{2}

in a closed vessel of one litre capacity. If 50%

N_{2}

is convened into

N H_{3}

at equilibriumm, the value

K_{c}

for the reaction is:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

2

moles of

N_{2}

are mixed with

6

moles of

H_{2}

in closed vessel of one litre capacity. If

50

N_{2}

is converted into

{N H}_{3}

at equilibrium, the value of

K_{c}

for the reaction,

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

Q. 2mol of

N_{2}

is mixed with 6 mol of

H_{2}

in a closed vessel of one litre capacity. If 50%

N_{2}

is converted into

N H_{3}

at equilibrium, the value of

K_{c}

for the reaction.

N_{2} (g) 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

is:

2

N_{2}

is mixed with

6

H_{2}

in a closed vessel of one litre capacity. If 50% of

N_{2}

is converted into

{N H}_{3}

at equiibrium, the value of

K_{c}

for the reaction ?

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

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2 mol of N2 is mixed with 6 mol of H2 in a closed vessel of one litre capacity. If 50% N2 is converted into NH3 at equilibrium, the value of KC for the reaction N2(g)+3H2(g)⇌2NH3(g) is:

The correct option is A 4/27 N2(g)+3H2(g)⇌2NH3(g)at t=0 2mol 6mol −at equilibrium 50% N2 is converted into NH3.at equim, (2-1) 3 2KC=[NH3]2[H2]3[N2]=(2)2(3)3×1=427

$2$ mol of $N_{2}$ is mixed with $6$ mol of $H_{2}$ in a closed vessel of one litre capacity. If $50$ % $N_{2}$ is converted into $N H_{3}$ at equilibrium, the value of $K_{C}$ for the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ is:

The correct option is A $4 / 27$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
at $t = 0$ 2mol 6mol $-$
at equilibrium 50% $N_{2}$ is converted into ${N H}_{3}$ .
at equim, (2-1) 3 2
$K_{C} = \frac{{[{N H}_{3}]}^{2}}{{[H_{2}]}^{3} [N_{2}]} = \frac{{(2)}^{2}}{{(3)}^{3} \times 1} = \frac{4}{27}$