please I need the graphs to question 2 above drawn and also the calculations. th

ID: 560867 • Letter: P

Question

please I need the graphs to question 2 above drawn and also the calculations. this is my second time sending this question please draw the graphs. I want to see them.

Principles l ICA 2017-18 Due Date: 17 November 2017 1. Initial rate data at 298 K are listed in the table below for the r Experiment Initial (NH Initial [NO 1 Initial Rate of 0.24 0.12 0.12 0,10 0.10 0.15 NH (M/s) 7.2 10 36 × 106 5.4 × 10 (a) What is the rate law? (b) What is the value of the rate constant ? (c) What is the initial rate when the initial concentrations are (NH41 0.39 M 12 marks] 12 marks) and [NO2 ] 0.052 M ? [3 marks] (a) At elevated temperatures, nitrous oxide decomposes according to the equation 2N20(g) 2 N2(g) + 02(g) 120 180 Time (min) 0 [N201/M 0.250 60 0.204 0.190 0.166 0.218 Using the above data, plot the appropriate graphs to determine whether the reaction is 1t order or 2nd order. What is the value of the rate constant? [10 marks Rate constants for the decomposition of azomethane CHMCHs(g) 2 CH3. (g) + N2(8) exhibit the following temperature dependence (b) 303 287 Temp/°C 250 k/s| |1.8 × 106 1.6×104 1.5 x10s 0x105-11.6x10 6.0 x 10 Plot an appropriate graph of the data and determine the activation energy for this reaction. 8 marks] A typical aspirin tablet contains 324 mg of aspirin (acety-salicylic acid, C,H O:) a monoprotic acid having K, 3.0 x 104. If you dissolve two aspirin tablets in a 300 mL glass of water, what is the pH of the solution and the percent dissociation? 8 marks]

Explanation / Answer

Compare with the linear regression equation and get

-Ea/R = -(25308 K)

====> Ea = R*(25308 K) = (8.314 J/mol.K)*(25308 K)*(1 kJ/1000 J) = 210.4107 kJ/ mol 210.41 kJ/mol.

2a) Start with the generalized equations for a 1st order and a 2nd order reaction. The rate equations are

1st. order:

ln [A] = ln [A]0 – k*t where [A] = concentration of the reactant at time t; [A]0 = concentration of the reactant at t =0 and k = 1st. order rate constant.

2nd. order:

1/[A] = /1[A]0 + k*t where k = 2nd. order rate constant.

Prepare the following table.

Time (min)

[N2O5] (M)

ln [N2O5]

1/[N2O5] (M-1)

0.250

-1.3863

4.0000

0.218

-1.5233

4.5871

0.204

-1.5896

4.9020

120

0.190

-1.6607

5.2631

180

0.166

-1.7958

6.0241

Plot ln [N2O5] vs t and 1/[N2O5] vs t in two separate plots.

Both the plots are linear; however, the plot of ln [N2O5] vs t has a R2 value closer to 1.000 and hence, the reaction is 1st order.

The linear equation for the 1st order equation is y = -0.0023x – 1.3863; compare this with the 1st order rate law to get the numerical value of k as 0.0023; the unit of k is s-1 and hence, the rate constant is 0.0023 s-1 (ans).

b) Use the Arrhenius equation. The exponential form of the Arrhenius equation is given as

ln k = ln A – Ea/RT where Es is the activation energy for the reaction, R is the gas constant and T is the absolute temperature; A is the pre-exponential factor.

Prepare a table of ln k vs T values as below.

Temperature (°C)

Temperature (K) = Temperature (°C) + 273

1/T (K-1)

k (s-1)

ln k

250

523

0.001912

1.8*10-6

-13.2277

268

541

0.001848

1.5*10-5

-11.1075

287

560

0.001786

6.0*10-5

-9.7212

303

576

0.001736

1.6*10-4

-8.7403

Plot ln k against 1/T as below.