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Standard IX

Chemistry

Temperature and Rate of Reaction

At the start ...

Question

At the start of a reaction, there are 0.249 mol3.21 x 10-2 mol H2 and 6.42 x 10-4 mol NH3.50 L reaction vessel at 375° C. Hence reactionwill proceedin(1) Forward direction (2) Backward direction(3) At equilibrium(4) Stops

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Similar questions

N_{2} + 3 H_{2} ⇌ 2 N H_{3} : K_{c} = 1.2

At the start of a reaction, there are

0.249 m o l N_{2}

3.21 \times 10^{- 2} m o l H_{2}

and

6.42 \times 10^{- 4} m o l N H_{3}

in a

3.50 L

reaction vessel at

375^{o} C

. Hence reaction will proceed in ?

Q. The equilibrium constant for the following reaction is 49 at

450^{o} C .

0.22 m o l

I_{2}

0.22 m o l

H_{2}

and

0.66 m o l

H I

were put in a evacuated 1 litre container.

H_{2} + I_{2} ⇌ 2 H I

Then, the correct statement will be:

Q. At

1400 K

, for the reaction

C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)

K_{c} = 2.5 \times 10^{- 3}

. A

10 L

reaction vessel contains 2 moles of

C H_{4},

3 moles of

C S_{2}

, 3 moles of

H_{2}

and 4 moles of

H_{2} S

. Choose the correct statement from the following:

Q. The value of

K_{c}

for the reaction

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

1 \times 10^{- 4}

. At a given time, the composition of reaction mixture is

[H I] = 2 \times 10^{- 5}

mol and

[I_{2}] = 1 \times 10^{- 5}

mol. The direction in which the reaction will proceed is (For forward direction write 0 and for reverse direction write 1)

A mixture of 1.57 mol of $N_{2}$ , 1.92 mol of $H_{2}$ and 8.13 mol of $N H_{3}$ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, $K_{c}$ for the reaction
$N_{2} (g) + 3 H_{2} (g) \leftrightarrow 2 N H_{3} (g) i s 1.7 \times 10^{2}$
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

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