Concentration Models Describe Changing Mixtures
A chemical process can change how much of a substance is present in a fixed volume. A concentration-versus-time model describes that change: it gives the concentration at each time. Its derivative answers a different question: how quickly is the concentration changing at a particular instant?
This distinction matters in practical interpretation. A concentration tells you the amount per unit volume at a time; a derivative tells you the instantaneous rate at which that amount per unit volume is increasing or decreasing. The sign of the derivative indicates the direction of change, while its units record how concentration changes per unit time.
Signs and Units Carry Meaning
Suppose concentration is measured in moles per litre and time in minutes. Then \(C(t)\) has units \(\mathrm{mol/L}\), and \(C'(t)\) has units \(\mathrm{mol/(L\cdot min)}\). The derivative is not itself a concentration. It measures how many moles per litre the concentration changes by, per minute, near the specified time.
The sign is part of the mathematical answer. If \(C'(t)=0.04\ \mathrm{mol/(L\cdot min)}\), the concentration is increasing at that instant: the substance is being formed, or its concentration is otherwise rising. If \(C'(t)=-0.04\ \mathrm{mol/(L\cdot min)}\), the concentration is decreasing at that instant: the substance is being depleted. The positive magnitude of that depletion is \(0.04\ \mathrm{mol/(L\cdot min)}\), but the signed derivative remains negative.
A derivative describes the change at one instant, not necessarily the change throughout an entire time interval. If the derivative varies with time, a rate at one time cannot be used as though it were the rate at every other time.
Worked Example: A Reactant Being Depleted
Worked Example: A Reactant Being Depleted
A reactant has concentration model \(C_A(t)=0.60-0.10t+0.005t^2\), where \(t\) is measured in minutes and concentration is measured in \(\mathrm{mol/L}\). Find the instantaneous rate of concentration change at \(t=4\) minutes, and interpret it.
For this model, compute the derivative using the difference quotient. At a general time \(t\), the change in concentration over an additional time \(h\) is
Since \((t+h)^2-t^2=2th+h^2\), divide by \(h\), with \(h\ne 0\):
The instantaneous rate is the limit of these average rates as \(h\) approaches zero. Therefore, \(C_A'(t)=-0.10+0.01t\). At \(t=4\),
The negative sign means the reactant concentration is decreasing at that instant. Its depletion magnitude is \(0.06\ \mathrm{mol/(L\cdot min)}\).
Check: The concentration at four minutes is \(C_A(4)=0.60-0.10(4)+0.005(4^2)=0.60-0.40+0.08=0.28\ \mathrm{mol/L}\). This is the concentration, not the rate. Differentiating the model as shown gives \(C_A'(t)=-0.10+0.01t\), and substituting \(t=4\) gives \(-0.06\ \mathrm{mol/(L\cdot min)}\), as calculated.
Why the Difference Quotient Gives an Instantaneous Rate
For two nearby times \(t\) and \(t+h\), the quotient \(\frac{C(t+h)-C(t)}{h}\) measures average concentration change per unit time over that interval. It is the slope of the secant line through the two corresponding points on the concentration graph. Letting \(h\) approach zero makes the second time approach the first. When the limit exists, the secant slope approaches the tangent slope, which is the derivative \(C'(t)\).
This explains why a derivative is a rate at an instant rather than just a difference between two concentration readings. It also explains why changing concentration models can have different rates at different times. For the quadratic model in the example, the derivative depends on \(t\); it is not one fixed number.
Rates Can Change While Concentration Rises
A positive derivative does not mean that concentration is constant, and it does not say that the rate itself is increasing. It means only that concentration is increasing at the specified instant. The derivative's value can be positive while getting smaller, for example. A derivative of zero means the instantaneous concentration change is zero at that instant; it does not by itself say that concentration is zero or that no process is occurring.
Worked Example: A Product Forming at a Changing Rate
A product concentration is modelled by \(C_P(t)=0.10+0.08t-0.01t^2\), for \(0\le t\le4\) hours, with concentration in \(\mathrm{mol/L}\). Find the rate of concentration change at \(t=2\) hours and interpret it.
Using the derivative of the model, \(C_P'(t)=0.08-0.02t\). Substituting \(t=2\) gives
The positive sign means the product concentration is increasing at two hours, at \(0.04\ \mathrm{mol/(L\cdot h)}\). The rate is not the concentration at that time: \(C_P(2)=0.10+0.08(2)-0.01(2^2)=0.10+0.16-0.04=0.22\ \mathrm{mol/L}\).
Check: Differentiating the stated model gives \(0.08-0.02t\); substituting two gives \(0.04\), confirming the rate. Substituting two into the original model gives \(0.22\ \mathrm{mol/L}\), confirming the separate concentration value. The rate has concentration-per-hour units, while the concentration has concentration units.
Time Units Must Match the Reported Rate
The time variable determines the time part of the derivative's units. A model written with \(t\) in minutes gives a rate per minute. If the requested rate is per second, convert the units rather than silently changing what \(t\) means. Since one minute is sixty seconds, a rate per minute is divided by sixty to express the same change per second.
Worked Example: Converting a Depletion Rate to Per Second
A concentration model is \(C(t)=0.80-0.12t\), where \(t\) is in minutes and \(C\) is in \(\mathrm{mol/L}\). Find the rate at \(t=3.5\) minutes, express it per second, and state the concentration at that time.
The model is linear, so its rate of change is its constant slope: \(C'(t)=-0.12\ \mathrm{mol/(L\cdot min)}\). Thus at \(t=3.5\) minutes, the signed rate is also \(-0.12\ \mathrm{mol/(L\cdot min)}\). To convert to a per-second rate, divide by \(60\):
The negative sign indicates depletion. The positive depletion magnitude is \(0.002\ \mathrm{mol/(L\cdot s)}\).
The concentration at \(3.5\) minutes is \(C(3.5)=0.80-0.12(3.5)=0.80-0.42=0.38\ \mathrm{mol/L}\).
Check: The derivative of \(0.80-0.12t\) is the constant \(-0.12\), so evaluating at \(3.5\) gives the stated per-minute rate. Multiplying the converted rate by \(60\ \mathrm{s/min}\) recovers \(-0.12\ \mathrm{mol/(L\cdot min)}\). Substitution into the original model gives \(0.38\ \mathrm{mol/L}\), a concentration value distinct from either rate.
A Common Sign and Quantity Mix-Up
One frequent error is to report the concentration itself as the rate. In the first example, the calculation \(C_A(4)=0.28\ \mathrm{mol/L}\) is correct, but it does not answer how quickly the concentration is changing. The requested rate is \(C_A'(4)=-0.06\ \mathrm{mol/(L\cdot min)}\). The function value and derivative value answer different questions and have different units.
A second error is to discard the negative sign and report the derivative as positive. For example, replacing \(C_A'(4)=-0.06\) with \(+0.06\) as the signed rate would reverse the meaning: it would say the concentration is increasing. It is correct to say that the magnitude of depletion is \(0.06\ \mathrm{mol/(L\cdot min)}\), but the signed rate of concentration change is \(-0.06\ \mathrm{mol/(L\cdot min)}\).
A Reliable Interpretation Process
When a concentration model and a time are given, keep the numerical calculation and the physical interpretation distinct.
Confirm which substance the model describes and the units used for concentration and time.
Use the derivative of the model and evaluate it at the specified time.
A positive value indicates increasing concentration; a negative value indicates depletion.
Keep concentration-per-time units with the signed derivative. If giving a positive depletion magnitude, identify it as a magnitude.
This process prevents three different ideas from being confused: how much substance is present, whether its concentration is increasing or decreasing, and how quickly that change occurs. In chemical applications, each is useful, but only the derivative supplies the instantaneous rate of concentration change.
Check Your Understanding
Answer each question with attention to the sign and units.
- If \(C'(t)=0.025\ \mathrm{mol/(L\cdot min)}\) at a specified time, is the substance being formed or depleted at that instant?
- A model gives \(C(5)=0.40\ \mathrm{mol/L}\) and \(C'(5)=-0.03\ \mathrm{mol/(L\cdot min)}\). Which value is the concentration, and which is the signed rate?
- If a signed derivative is \(-0.08\ \mathrm{mol/(L\cdot h)}\), what is the positive magnitude of depletion?
- A constant rate is \(-0.18\ \mathrm{mol/(L\cdot min)}\). Express it in \(\mathrm{mol/(L\cdot s)}\).
- What does a derivative of zero at one instant tell you, and what does it not establish about the concentration?
