The rate constant for the first-order reaction- 2 N 2 O 5- 4 N O 2- O 2 is 3-0 - 10-5 s -1- If the rate is 2-40 - 10-5 mol - L s -1- then the initialconcentration of N 2 O 5 in - mol - L isA- 1-2B- 0-8C- 0-04D- 1-4

The correct option is B 0.8 Rate = K[N_2O_5 ] 2.40 × 10^ 5 = 3.0 × 10^ 5 × [N_2O_5] [N_2O_5] = 2.40 ×10^ 5/3.0 ×10^ 5 = 0.8 mol/L ...

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Byju's Answer

Standard XII

Chemistry

Zero Order Reaction

The rate cons...

Question

The rate constant for the first-order reaction, 2 $N_{2}$ $O_{5}$ $\to$ 4 ${N O}_{2}$ + $O_{2}$ is 3.0 $\times$ $10^{- 5}$ $s^{- 1}$ . If the rate is 2.40 $\times$ $10^{- 5}$ $\frac{m o l}{L}$ $s^{- 1}$ , then the initial concentration of $N_{2}$ $O_{5}$ (in $\frac{m o l}{L}$ ) is

1.4

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1.2

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0.04

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0.8

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Solution

The correct option is D
0.8

Rate = K[ $N_{2} O_{5}$ ]

2.40 $\times$ $10^{- 5}$ = 3.0 $\times$ $10^{- 5}$ $\times$ [ $N_{2}$ $O_{5}$ ]

[ $N_{2}$ $O_{5}$ ] = $\frac{2.40 \times 10^{- 5}}{3.0 \times 10^{- 5}}$ = 0.8 $\frac{m o l}{L}$

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Similar questions

Q. For the reaction,

2 N_{2} O_{5} ⇌ 4 N O_{2} + O_{2}

rate and rate constant are

2.40 \times 10^{- 5}

mol

L^{- 1} s^{- 1}

and

3 \times 10^{- 5} s^{- 1}

respectively, then calculate the concentration of

N_{2} O_{5}

Q. The rate constant for the reaction

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

3.0 \times 10^{- 5} {sec}^{- 1}

. If the rate is

2.4 \times 10^{- 5} m o l L^{- 1} {sec}^{- 1}

, then the concentration of

N_{2} O_{5} (i n m o l e L^{- 1}

) is

Q. The rate constant for the reaction,

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

, is

3.0 \times 10^{- 5} s e c^{- 1}

. If the rate is

2.40 \times 10^{- 5}

mol

l i t r e^{- 1} s e c^{- 1}

, then the concentration of

N_{2} O_{5}

( in mol

l i t r e^{- 1}

) is:

Q. The rate constant for the reaction,

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

2 \times 10^{- 5} s^{- 1}

. If rate of reaction is

1.4 \times 10^{- 5} m o l L^{- 1} s^{- 1}

, what will be the concentration of

N_{2} O_{5}

in mol

L^{- 1}

Q. For the reaction,

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

rate and rate constant are

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

and

3.4 \times 10^{- 5} s^{- 1}

respectively. The concentration of

N_{2} O_{5}

in mol

L^{- 1}

will be:

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The rate constant for the first-order reaction, 2N2O5 → 4NO2 + O2 is 3.0 × 10−5 s−1. If the rate is 2.40 × 10−5 molL s−1, then the initial concentration of N2O5 (in molL) is

The correct option is D 0.8 Rate = K[N2O5 ] 2.40 × 10−5 = 3.0 × 10−5 × [N2O5] [N2O5] = 2.40× 10−53.0× 10−5 = 0.8 molL

The rate constant for the first-order reaction, 2 $N_{2}$ $O_{5}$ $\to$ 4 ${N O}_{2}$ + $O_{2}$ is 3.0 $\times$ $10^{- 5}$ $s^{- 1}$ . If the rate is 2.40 $\times$ $10^{- 5}$ $\frac{m o l}{L}$ $s^{- 1}$ , then the initial concentration of $N_{2}$ $O_{5}$ (in $\frac{m o l}{L}$ ) is

The correct option is D
0.8

Rate = K[ $N_{2} O_{5}$ ]

2.40 $\times$ $10^{- 5}$ = 3.0 $\times$ $10^{- 5}$ $\times$ [ $N_{2}$ $O_{5}$ ]

[ $N_{2}$ $O_{5}$ ] = $\frac{2.40 \times 10^{- 5}}{3.0 \times 10^{- 5}}$ = 0.8 $\frac{m o l}{L}$