In the following reaction 2H2O2- 2H2O-O2 rate of formation of O2 is 3-6 M min-1- The rate of formation of H2O is-

The activation energy of the reaction:
$A + B \to$ Products is 105.73 KJ/mol. At $40^{\circ} C$ , the products are formed at the rate of 0.133 mol $L^{- 1} m i n^{- 1}$ the rate of formation of products at $80^{\circ} C$ will be

Q. If the rate of decomposition of

N_{2} O_{5}

during a certain time internal is

2.4 \times 10^{- 4} m o l L^{- 1} m i n^{- 1}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

What is the rate of formation of

N O_{2}

and

O_{2} m o l L^{- 1} m i n^{- 1}

In a reaction, A → Products. The concentration of A decreases from 0.5 mol L^-1 to 0.3 mol L^-1 in 10 min. The rate during this interval will be

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In the following reaction 2H2O2→2H2O+O2 rate of formation of O2 is 3.6 M min−1. The rate of formation of H2O is:

The correct option is C 7.2mollitre−1min−1The rate of formation of water is twice the rate of formation of oxygen.d[H2O]dt=2d[H2O]dt=2×3.6Mmin−1=7.2mollitre−1min−1

In the following reaction $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ rate of formation of $O_{2}$ is 3.6 M $m i n^{- 1} .$ The rate of formation of $H_{2} O$ is:

The correct option is C $7.2 m o l l i t r e^{- 1} m i n^{- 1}$
The rate of formation of water is twice the rate of formation of oxygen.
$\frac{d [H_{2} O]}{d t} = 2 \frac{d [H_{2} O]}{d t} = 2 \times 3.6 M m i n^{- 1} = 7.2 m o l l i t r e^{- 1} m i n^{- 1}$