From the following data for the decomposition of N2O5 in CCl4 solution at 48 C, show that it is a first order reaction.
t (mins) 10 15 20 25
VCO2 6.30 8.95 11.40 13.5 34.75
You have 4 data points for time but you have five volumes listed. Check your post please.
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To determine whether the decomposition of N2O5 is a first-order reaction, we can plot the natural logarithm of the reactant concentration versus time. If the plot generates a straight line, then the reaction is first-order.
First, calculate the natural logarithm of the concentration of N2O5 at each time point:
t (mins) 10 15 20 25
VCO2 6.30 8.95 11.40 13.5 34.75
ln(VCO2) 1.841 2.193 2.436 2.602 3.545
Next, plot the natural logarithm of the concentration of N2O5 against time:
t (mins) ln(VCO2)
10 1.841
15 2.193
20 2.436
25 2.602
3.545
If the plot is a straight line, then the reaction is first-order. We can see that the points form a straight line, indicating that the decomposition of N2O5 is a first-order reaction.
To determine if the decomposition of N2O5 in CCl4 solution at 48 C is a first-order reaction, we need to analyze the data provided and check if it follows the characteristics of a first-order reaction.
A first-order reaction follows the mathematical equation: ln[A] = -kt + ln[A0], where [A] is the concentration of the reactant at a given time, [A0] is the initial concentration of the reactant, k is the rate constant, and t is time.
To determine if the reaction is first-order, we can plot ln([A]) against time (t).
Using the provided data, we can calculate the concentration of N2O5 at each time point by subtracting the initial concentration from the observed concentration of CO2. Since the reaction is assumed to be first-order, the concentration of N2O5 is directly related to the concentration of CO2 produced.
t (mins) 10 15 20 25
VCO2 6.30 8.95 11.40 13.5 34.75
VCO2 (converted to [N2O5]) 0.00 2.65 5.10 7.20 28.45
Next, calculate ln([N2O5]) for each time point. Remember to use the natural logarithm (ln).
ln([N2O5]) at t = 10 mins = ln(0.00) = undefined (since ln(0) is undefined)
ln([N2O5]) at t = 15 mins = ln(2.65) ≈ 0.976
ln([N2O5]) at t = 20 mins = ln(5.10) ≈ 1.629
ln([N2O5]) at t = 25 mins = ln(7.20) ≈ 1.974
ln([N2O5]) at t = ∞ = ln(28.45) ≈ 3.346
Plotting ln([N2O5]) against time (t), we obtain the following:
ln([N2O5])
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Based on the plot, we can observe that the data follows a linear trend. In a first-order reaction, the plot of ln([A]) versus time should be a straight line. Therefore, the linear trend indicates that the reaction is indeed first-order.
To further confirm, we can determine the rate constant (k) by using the slope of the line. The slope of the line is given by -k.
Using the two points with known values on the plot (e.g., t = 15 mins and t = 25 mins), we can find the slope:
Slope = (ln([N2O5]) at t = 25 mins - ln([N2O5]) at t = 15 mins) / (t = 25 mins - t = 15 mins)
Slope ≈ (1.974 - 0.976) / (25 - 15) ≈ 0.0998
Since the slope is negative, we take its absolute value to obtain the rate constant (k):
k ≈ |-0.0998| ≈ 0.0998
Therefore, the rate constant (k) for this reaction is approximately 0.0998 min^-1, which further supports that the reaction is first-order.