Question

a)A concentration C(mol/L) varies with time (min) according to the equation C=3.00exp(−2.00t) a) What are the implicit units of 3.00 and 2.00? b) Suppose the concentration is measured at t = 0 and t = 1 min. Use two-point linear interpolation or extrapolation to estimate C(t=0.6min) and t(C=0.10mol/L) from the measured values, and compare these results with the true values of these quantities. c) Sketch a curve of C versus t, and show graphically the points you determined in Part (b).

Answer 1

Answer:

a. 3.00 must be in mol/L and 2.00 in 1/min.

b. Using lineal equation:

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

Using exponential equation:

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

Explanation:

a) C is in mol/L so, as we know that the exponential does not have units, 3.00 must be in mol/L and 2.00 in 1/min.

b) If we want to make a line between 0 an 1 we need to find the linear equation for this.

We can use the slope intercept-equation:

y = ax + b (1)

Let's find a and b.

a is the slope of here, so we can use the next equation to find it:

$a=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

We have two points, (0,C₁) and (1,C₂)

$C_{1}(0)=3.00e^{-2.00*(0)}=3.00 mol/L$

$C_{2}(1)=3.00e^{-2.00*(1)}=0.41 mol/L$

So a will be:

$a=\frac{0.41-3.00}{1-0}=-2.59$

We can use the equation (1) to find b, for instance, let's choose: $y_{1}=ax_{1}+b$

$b=y_{1}-ax_{1}=C_{1}-(2.59*0)=3.00$

Finally, our linear equation will be: $y=-2.59x+3.00$ or  $C=-2.59t+3.00$

Using lineal equation:

t = 0.6 min then C = 1.45 mol/L

C = 0.10 mol/L then t = 1.12 min

Using exponential equation:

t = 0.6 min then C = 0.90 mol/L

C = 0.10 mol/L then t = 1.70 min

I hope it helps you!