zero order reaction

[FrankieClarke2C]

Why is it a zero order reaction with slope -k if the straight line fits to the data?

[Briana Lopez 4K]

The slope is -k because the integrated rate law for zero order is
-kt + [A]o = [A]

similar to the straight line y= mx + b

[A] = -kt + [A]o

[Aili Ye 4L]

The integrated rate for a first order reaction is A= -kt + Ao which is in the form of a linear equation where -k is the slope.

[Chantel_2I]

A zero order equation integrated gives a first order equation, which when graphed, gives a straight line, as opposed to a second order equation, like something with x^2, that gives a parabola.

[Jennifer Fuentes 2K]

because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of k, a positive value.

[Kimia Rategh 2A]

Hi! Zero order reaction rates are independent on reaction concentrations, therefore we would have a reaction rate similar to y= mx +b where -k is the slope and the initial concentration is the intercept. Therefore, if the reaction is a zero order, then the data should fit the line with a slope of -k. Hope this helps!

[Tammy Shen 2L]

A zero order reaction has the integrated rate law A=-kt+A0, which mimics the linear equation y=mx+b. This means that -k acts as the slope.

[Zoe Dhalla 3I]

Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. For a zero order reaction, as shown in the following figure, the plot of [A] versus time is a straight line with k = - slope of the line.

[Heba Bounar 3K]

Hi! The integrated rate law of a zero order reaction is [A} = -kt +[A]o. This resembles the linear equation y = mx+b in which m represents slope. Therefore, -k represents the slope. Therefore, if a straight line is fit to the data ([A] vs time), then we know it is zero order and the slope is -k. Hope that helps!

[Joellen 1B]

we know that the zero order reaction equation is A=-kt+A0, and it is similar to the equation y = mx +b where m represents the slope, so k would represent the slope and it is negative in this case.

[Andrewtrung Le 3A]

With a zero-order reaction, you've already recognized that the straight line fits to the data. Any line that is straight has a constant slope. Also, the slope is -k because k measures the rate at which reactants are used up - never can reactants increase in concentration when a chemical reaction progresses!

[Andrewtrung Le 3A]

With a zero-order reaction, you've already recognized that the straight line fits to the data. Any line that is straight has a constant slope. Also, the slope is -k because k measures the rate at which reactants are used up - never can reactants increase in concentration when a chemical reaction progresses!

[Jeffrey Vo 2A]

Hi Frankie, you already know that a zero order reaction has the equation of A=-kt+A0. This integrated rate law of a zero order reaction is akin to y=mx+b, where you can probably see that -k is the slope of A=-kt+A0. On a more conceptual level, the slope being negative is a result of the rate how how reactants are becoming products, so you can think about it being negative as there is a loss in reactants naturally in a chemical reaction just like how the concentration of products would have the value of k being positive.

[Alice Weber 3I]

Hi, we would use this equation A=-kt+A0 which is similar to y=mx+b, therefore, -k is the slope.

[Ashley Presnell 1C]

This is because it is showing the concentration over time of the reaction, and since its linear it means the reagent is decreasing linearly over time which agrees with the zero order rules.

[Jieun 2C]

The slope is -k because A= -kt + Ao is in the form of y=mx+b where m=-k and m is the slope.

[Kelly McFarlane]

The reaction rate of a zero order reaction is [A]=-kt + [A]0 so following the rule y=mx+b, y is the y-axis ([A]), x is the x-axis (t), b is the y intercept ([A]0), and m is the slope (-k).

[Nishan Reddy 3K]

This is just the common way to represent the first order reaction where ln[A] = -kt + A0. This means that the reaction rate isn’t directly proportional to time.

[Jacob_Eberson_2D]

the rate law for zero order is A=-kt+A0, which has the same format as the linear equation y=mx+b (with -k as the slope)

[Morgan Micallef 1A]

Hello! The intergrated equation for a zero-order reaction is A=-kt+A0. According to this equation, the slope of the line is -k

[Isabelle Rieke-Wey 2G]

Hi!
the integrated equation for a zero-order reaction is A=-kt+A0, similar to y=mx+b which is why the slope is -k.

[905740390]

The zero-order reaction is A = -kt +A0 which tells us that the slope of this reaction is -k if you view this equation similar to y = mx + b

[Konmal Ali 1G]

A graph representing the concentration of any reactant as a function of time is a straight line with a slope of k since rate is independent of reactant concentration. Because the concentration of the reactant falls with time, the value of k is negative.

[Aanjaneyaa]

Since, there is integration involved, it ends up being the zero order reaction becomes a first order equation. The zero order reaction is A=-kt+A0, and based on the equation y=mx+b, m=-k, so the slope is -k.

[Anish_Marripati_2F]

Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time

[605823283]

it is an intergrated rate law

[Neha Mukund]

When one observes the equation to the zero order rate law, A = -kt +A0, it can be seen that the slope is -k. Hope that helps.

[Alejandra Hernandez 2A]

Hi!! A reaction that's zero-order has the integrated rate law of A=-kt+A0, this law mimics the linear equation y=mx+b (or uses it as a model). This means that -k acts as the 'm' aka the slope. Hope this helped!! :)

[gwenkelley3L]

since the integrated equation for a zero order reaction is A = -kt+A0, the slope is -k since this equation is very similar y = mx+b

[Sophia Hartwell 1F]

Hi there! So because there is integration involved, the zero order reaction becomes a first order equation. The zero order reaction is A=-kt+A0, and the slope is -k based on the equation y=mx+b, m=-k.