[A] v. Time

[LNgo 1G]

For a first order reaction the [A] v Time plot is a decreasing exponential but what would the graph of [A] v Time look like for a second order reaction?

[Haley Dveirin 1E]

For a second order reaction it gives a straight line.

[Alfred Barrion 2H]

The graph of [A] v. Time would be linear.

[Minh Ngo 4G]

Linear

[Angela Patel 2J]

Would the slope of the graph be k?

[805303639]

The graph of 1/[A] v Time for a second-order reaction yields a line with slope=k and y-intercept=1/[A]i. Hence the graph of [A] v Time for a second-order reaction would not be linear. The graph would be a rational function with a vertical asymptote at t= -1/(k[A]i).

[nicole-2B]

Would the slope of the graph be k?

yes the slope would be k

[Shrayes Raman]

Linear 1/[A] v time but not [A] v time

[Brandon Tao 1K]

The second order is linear

[Jarrett Peyrefitte 2K]

It would be linear

[Nawal Dandachi 1G]

for second order, it should be linear

[JesseAuLec1Dis1G]

It's linear for second order. Does anyone know when it wouldn't be linear?

[Caroline Zepecki]

For a second order reaction, it would be linear.

[McKenna_4A]

It would be linear, confirming that it is a 2nd order rxn.

[Emily Vainberg 1D]

If you look at the graoh of 1/[A] v Time for a second-order reaction, the slope is equal to K. The y intercept =1/[A]i. This means the graph for a second order reaction would be linear.

[Jaci Glassick 2G]

The slope of a 2nd order reaction (1/[A] vs. time) is k. The graph is linear.

[Abigail Sanders 1E]

The graph of [A] vs time would not be linear for a second order reaction. It is only linear for zero order. For second order the graph of 1/[A] vs t is a linear function with a slope equivalen to +k.

[Jacob_Eberson_2D]

it would be linear with a positive slope (k)

[Saatvika Nair]

I believe that the second order would be a linear function (straight line) with a positive slope of +k. The graph would be 1/[A]. Hope that helps!

[205769933]

Every graph should show a linear graph, it's the axis that are different.

[JennyZhu1K]

ln[A] vs time will be a straight line for first order reactions and [A] vs time will be a straight line for zero order reactions.

[Ryan Khiev 1L]

[A] vs time for a second order reaction would look like a steeper exponential function than a decreasing exponential, it could only be linear if it was 1/[A] vs time. If you have a graphing calculator and graph 1/X it would look similar to that.

[Emma Goellner 2I]

It should be a linear graph for a second order reaction!

[905756606]

For a second order reaction, the graph of 1/[A] (not [A]) vs time would be linear with a slope = k.

[Michelle Gong]

It would be linear with a positive K+ slope!

[Kelly McFarlane]

The plot of a second order reaction on a [A] vs. time graph would be linear.

[Violet Mbela 2B]

Hi,
The plot of a second-order reaction would be a linear (straight line) graph with a positive slope of K+.

[almaortega]

Second order reactions when plotted 1/[A] vs time will be an increasing linear graph. k will be the slope of this line.