The rates of a certain reaction (dc/dt) at di | vedclass

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(A) The rate of a reaction is defined as the change in concentration of reactants or products per unit time. For a zero-order reaction,the rate of rea

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Zero order

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(A) The rate of a reaction is defined as the change in concentration of reactants or products per unit time. For a zero-order reaction,the rate of reaction is independent of the concentration of the reactants,i.e.,$	ext{Rate} = k[A]^0 = k$. In the given data,the rate of the reaction remains approximately constant $(2.8 	imes 10^{-2} \ mole \ litre^{-1} \ sec^{-1})$ at all given time intervals $(0, 10, 20, 30 \ sec)$. Since the rate does not change with time,the reaction must be of zero order.

Time $(sec)$	Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$	$2.8 \times 10^{-2}$
$10$	$2.78 \times 10^{-2}$
$20$	$2.81 \times 10^{-2}$
$30$	$2.79 \times 10^{-2}$

Time $(s)$	Concentration of the reactant $(mol \ L^{-1})$
$0$	$0.1$
$15$	$0.05$
$20$	$0.1 - x$

The rates of a certain reaction $(dc/dt)$ at different times are as follows:
Time $(sec)$ Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$ $2.8 \times 10^{-2}$
$10$ $2.78 \times 10^{-2}$
$20$ $2.81 \times 10^{-2}$
$30$ $2.79 \times 10^{-2}$

The reaction is:

Similar Questions

For the reaction $A \longrightarrow 3 B$,the rate is given by $rate = k[A]^0$. Based on the data provided in the table,what is the concentration of the product $B$ (in $mol \ L^{-1}$) after $20 \ s$?
Time $(s)$ Concentration of the reactant $(mol \ L^{-1})$
$0$ $0.1$
$15$ $0.05$
$20$ $0.1 - x$

Decomposition of $X$ exhibits a rate constant of $0.05 \ \mu g/year$. How many years are required for the decomposition of $5 \ \mu g$ of $X$ into $2.5 \ \mu g$?

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The decomposition of $NH_{3}$ on a platinum surface is a zero-order reaction. What are the rates of production of $N_{2}$ and $H_{2}$ if $k = 2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$?

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The rates of a certain reaction $(dc/dt)$ at different times are as follows:Time $(sec)$Rate $(mole \ litre^{-1} \ sec^{-1})$$0$$2.8 \times 10^{-2}$$10$$2.78 \times 10^{-2}$$20$$2.81 \times 10^{-2}$$30$$2.79 \times 10^{-2}$The reaction is: