The graph drawn between rate of reaction $V s$ concentration of reactant, gives a straight line through origin. The slop is $2.4 \times 10^{- 4} {min}^{- 1}$
Order of reaction is ?

zero

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one

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two

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three

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Solution

The correct option is A one
Since the slope of straight line $= 2.4 \times 10^{- 4} m i n^{- 1}$
unit shows that reaction is of $1$ order.
Option B is correct

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The graph drawn between rate of reaction Vs concentration of reactant, gives a straight line through origin. The slop is 2.4×10−4min−1 Order of reaction is ?

The correct option is A oneSince the slope of straight line =2.4×10−4min−1unit shows that reaction is of 1 order.Option B is correct

The graph drawn between rate of reaction $V s$ concentration of reactant, gives a straight line through origin. The slop is $2.4 \times 10^{- 4} {min}^{- 1}$
Order of reaction is ?

The correct option is A one
Since the slope of straight line $= 2.4 \times 10^{- 4} m i n^{- 1}$
unit shows that reaction is of $1$ order.
Option B is correct