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Moved the primary equation into its own line
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Avogadro ♦
Avogadro ♦
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The equation that you have derived is correct. For a zeroth-order reaction, $\log t_{1/2}=\log a-\log2K$.
$$\log t_{1/2}=\log a-\log2K$$ By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

The equation that you have derived is correct. For a zeroth-order reaction, $\log t_{1/2}=\log a-\log2K$.
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

The equation that you have derived is correct. For a zeroth-order reaction, $$\log t_{1/2}=\log a-\log2K$$ By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

typography corrected
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edit approved Jan 30, 2022 at 21:51
Loong
edit approved Jan 30, 2022 at 21:51
Loong
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The equation that you have derived is correct. For a zeroth order-order reaction  , $logt_{1/2} = loga - log2K$$\log t_{1/2}=\log a-\log2K$.
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the K$K$ value is very small. IncaseIn case of any reaction having K < 0.5$K\lt0.5$, 2K$2K$ becomes less than 1$1$ and hence log2K$\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero order-order reaction having a small Rate Constantrate constant (less than 0.5$0.5$).

The equation that you have derived is correct. For a zeroth order reaction  , $logt_{1/2} = loga - log2K$.
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the K value is very small. Incase of any reaction having K < 0.5, 2K becomes less than 1 and hence log2K becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero order reaction having a small Rate Constant(less than 0.5).

The equation that you have derived is correct. For a zeroth-order reaction, $\log t_{1/2}=\log a-\log2K$.
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

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Krish Ratta
Krish Ratta
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The equation that you have derived is correct. For a zeroth order reaction , $logt_{1/2} = loga - log2K$.
By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the K value is very small. Incase of any reaction having K < 0.5, 2K becomes less than 1 and hence log2K becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero order reaction having a small Rate Constant(less than 0.5).