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Factors Affecting Rate of Reaction

What is the u...

Question

What is the unit of Rate constant in case of the reaction $2 N O_{2}$ $+$ $F_{2}$ $=$ $2 N O_{2} F$ , where the law of mass action is expressed as Rate = $k$ $[N O_{2}]^{2}$ $[F_{2}]$ ?

(m o l)

l i t^{- 1}

t i m e^{- 1}

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m o l^{- 2}

l i t^{2}

t i m e^{- 1}

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(m o l)

l i t^{2}

t i m e^{- 1}

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None of these

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Solution

The correct option is B $m o l^{- 2}$ $l i t^{2}$ $t i m e^{- 1}$
Rate = $k$ $[N O_{2}]^{2}$ $[F_{2}]$ , The order of the reaction here is $(2 + 1)$ that is $3$ . The unit of rate is $(m o l)$ $l i t^{- 1}$ $t i m e^{- 1}$ .
And the unit of concentration is $(m o l)$ $l i t^{- 1}$ . Thus-
$(m o l)$ $l i t^{- 1}$ $t i m e^{- 1}$ $=$ $k$ $m o l^{3}$ $l i t^{- 3}$
From this, the unit of rate constant $k$ is $m o l^{- 2}$ $l i t^{2}$ $t i m e^{- 1}$

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Similar questions

2 N O_{2}

+

F_{2}

=

2 N O_{2} F

2 N O + O_{2} \to 2 N O_{2}

= k {[N O]}^{2} [O_{2}]

k = 2.0 \times 10^{- 6} {m o l}^{- 2} L^{2} s^{- 1}

[N O] = 0.04

m o l

L^{- 1}

[O_{2}] = 0.2

m o l

L^{- 1}

2 {N O}_{2} ⟶ N_{2} O_{2} + O_{2}

- \frac{d [{N O}_{2}]}{d t} = k {[{N O}_{2}]}^{n}

k = 3 \times 10^{- 3} {m o l}^{- 1} L {s e c}^{- 1}

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

{N O}_{2}

L^{- 1}

A ⟶ B

2 \times 10^{- 4}

l i t . {m o l}^{- 1} {m i n}^{- 1}

(1 / 12) \times {s e c}^{- 1}

2 N O_{2} F \to 2 N O_{2} + F_{2}

= K [N O_{2} F]^{2}

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