JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 9)
The given plots represent the variation of the
concentration of a reactant R with time for two
different reactions (i) and (ii). The respective
orders of the reactions are :
_9th_April_Morning_Slot_en_9_1.png)
0,1
1,0
0,2
1,1
Explanation
For 1st order reaction, we know
$$\ln {{{R_0}} \over {{R_T}}} = kt$$
$$ \Rightarrow $$ $$\ln {R_0}$$ - $$\ln {R_t}$$ = kt
$$ \Rightarrow $$ $$\ln {R_t}$$ = - kt + $$\ln {R_0}$$
So, $$\ln {R_t}$$ vs t is a straight line with negative slope.
$$ \therefore $$ First graph represent 1st order reaction.
For Zeroorder reaction, we know
R0 - Rt = kt
$$ \therefore $$ Rt = - kt + R0
So, $${R_t}$$ vs t is a straight line with negative slope.
$$ \therefore $$ Second graph represent zero order reaction.
$$\ln {{{R_0}} \over {{R_T}}} = kt$$
$$ \Rightarrow $$ $$\ln {R_0}$$ - $$\ln {R_t}$$ = kt
$$ \Rightarrow $$ $$\ln {R_t}$$ = - kt + $$\ln {R_0}$$
So, $$\ln {R_t}$$ vs t is a straight line with negative slope.
$$ \therefore $$ First graph represent 1st order reaction.
For Zeroorder reaction, we know
R0 - Rt = kt
$$ \therefore $$ Rt = - kt + R0
So, $${R_t}$$ vs t is a straight line with negative slope.
$$ \therefore $$ Second graph represent zero order reaction.
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