Question

Part I1 short qaestioes) 27 points (Write your answers clearty) I. Answer the following questions based on the phase diagram of a pure substance given in Fig. 1. A, B and C are three physical states O 2pt) Which of the following statements is/are correct? Statement E: The density of solid is greater than the density of liquid. Statemeat IH: By applying pressure, solid can be transformed into liquid a)Ib) eandII mone of the above. Temperature (i) Cpt) Which of the following statements ivare correct? Fig. I Statement I: Sublimation can occur at 1 atm. Statement II: Above the critical point, the gas can be compressed into liquid. a)I b)II c)I and II d) none of the above 2. An integrative rate law experiment at a given temperature shows that the plot of In [H:Oa] vs. time is a straight line (in Fig. 2) for the reaction: H2O2 (g) O2 (g) + H2O (g). [H-O (M) 2.600 2.064 1.638 1.300 1.032 0.819 Time (sec) 0.0000 400.00 800.00 1200.0 1600.0 2000.0 2400.0 0.650 (i) (2pt) The reaction order ofthis reaction is (ii) (2pt) Determine the value of the rate constant and its 00 800 1200 2000 2400 time (sec) unit. Fig. 2 (iii) (2pt) How much time does it take for HaO2 to decrease from 2.600 M to 0.325 M?

Answer 1

1. (i) Volume occupied by liquids is more than volume occupied by solids and

density = mass/Volume

Density is inversely proportional to volume. So, density of solids is greater than the density of liquids

But fro phase diagram we can not convert solid to liquid by applying pressure but reverse reaction is possible

so, answer is option a) (I)

(ii) Sublimation can takes place at 1 atm and at - 78^oC.So, yes its possible

But Critical point is a point above which a gas cannot be liquified

So, option a) I

2. (i)When we plot a graph between ln [H₂O₂[ and time, line is a straight line. thus, reaction is of first order.

(ii) For first order reactions, integrated rate aw is:

ln[H₂O₂] = ln [H₂O₂]_o - kt

Where k is the rate constant and be determined by the slope

From graph slope =(-0.2)-0.5 / (2000 - 800) = -5.83 *10^-4 s^-1

-k = -5.83 *10^-4 s^-1

k = rate constant = 5.83 *10^-4 s^-1

Where sec^-1 are the units

(iii) ln[H₂O₂]_o = ln ( 2.6) = 0.956

ln [H₂O₂] = ln (0.325) = -1.124

Now, ln[H₂O₂] = ln [H₂O₂]_o - kt

-1.124 = 0.956 - 5.83 *10^-4(t)

t = 3566 sec