For the reaction N2-3H2- 2NH3 if d-NH3-dt-2- 10-4 mol L-1s-1 the value of -d-H2-dt would be

From rate law nbsp; nbsp; nbsp; d[N_2]/dt= d[H_2]/3*dt= d[NH_3]/2*dt Rate of consumption of Nitrogen nbsp; = Rate of formation of Hydrogen / 3 = R ...

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Byju's Answer

Standard X

Chemistry

Rate of Reaction

For the react...

Question

For the reaction $N_{2} + 3 H_{2} \to 2 N H_{3} i f d [N H_{3}] / d t = 2 \times 10^{- 4} m o l L^{- 1} s^{- 1} the value of - d [H_{2}] / d t$ would be

1 \times 10^{- 4} m o l L^{- 4} s^{- 1}

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3 \times 10^{- 4} m o l L^{- 1} s^{- 1}

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4 \times 10^{- 4} m o l L^{- 1} s^{- 1}

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6 \times 10^{- 4} m o l L^{- 1} s^{- 1}

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Solution

The correct option is B $3 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
From rate law

$- d [N_{2}] / d t = - d [H_{2}] / 3 * d t = - d [N H_{3}] / 2 * d t$

Rate of consumption of Nitrogen
= Rate of formation of Hydrogen / 3
= Rate of formation of Ammonia / 2

Suggest Corrections

Similar questions

For the reaction:

N_{2} + 3 H_{2} \to 2 N H_{3}

\frac{d [N H_{3}]}{d t} = 2 \times 10^{- 4} m o l L^{- 1} s^{- 1}

, the value of

\frac{- d [H_{2}]}{d t}

would be:

Q. For the reaction,

N_{2} + 3 H_{2} ⟶ 2 N H_{3}

\frac{d [N H_{3}]}{d t} = 2 \times 10^{- 4} m o l L^{- 1} s^{- 1}

, the value of

\frac{d [H_{2}]}{d t}

would be:

Q. If

\frac{d [N H_{3}]}{d t} = 34 g h r^{- 1}

for the reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

then

\frac{d [H_{2}]}{d t}

is ______

g h r^{- 1}

Q. For the reaction

2 N H_{3} \to N_{2} + 3 H_{2}

- \frac{d [N H_{3}]}{d t} = k_{1} [N H_{3}], \frac{d [N_{2}]}{d t} = k_{2} [N H_{3}], \frac{d [H_{2}]}{d t} = k_{3} [N H_{3}]

then the relation between

k_{1}, k_{2}

and

k_{3}

is?

Q. For the reaction;

N_{2} + 3 H_{2} \to 2 N H_{3}

, the rate

\frac{d [N H_{3}]}{d t} = 2 \times 10^{- 4} M s^{- 1}

. Therefore, the rate

- \frac{d [N_{2}]}{d t}

is given as:

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For the reaction N2+3H2→2NH3 if d[NH3]/dt=2×10−4 mol L−1s−1 the value of −d[H2]/dt would be

The correct option is B 3×10−4 mol L−1s−1From rate law −d[N2]/dt=−d[H2]/3∗dt=−d[NH3]/2∗dt Rate of consumption of Nitrogen = Rate of formation of Hydrogen / 3 = Rate of formation of Ammonia / 2

For the reaction $N_{2} + 3 H_{2} \to 2 N H_{3} i f d [N H_{3}] / d t = 2 \times 10^{- 4} m o l L^{- 1} s^{- 1} the value of - d [H_{2}] / d t$ would be

The correct option is B $3 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
From rate law

$- d [N_{2}] / d t = - d [H_{2}] / 3 * d t = - d [N H_{3}] / 2 * d t$

Rate of consumption of Nitrogen
= Rate of formation of Hydrogen / 3
= Rate of formation of Ammonia / 2