Chemistry

4.31 To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A- (E_a/RT). equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/ (x₂-x₁)

By substituting the values, slope = -12.301

? –E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

? _aE= 8.314 JK^-1mol^-1 * 12.301 K

? _aE= 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

(? At T = 273K, ln k =-7.147)

On solving, we get ln A = 37.911

? A = 2.91*10⁶

When T = 30⁰C, hence T = 30 + 273 = 303K

? ln k =-2.8

? k = 6.08x10^-2s^-1.

When T = 50⁰C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 * 10⁶, E_a = 102.27 kJ mol^-1

Ln K = ln (2.91 * 10⁶)- 102.27/ (8.314 * 323)

Thus, ln k = -0.5 and k = 0.607s^-1.

4.30 When t = 0, the total partial pressure is P₀ = 0.5 atm

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2 (P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]_° -Initial concentration of reactant [R]-Concentration of reactant at time 't'

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

t= 2.303/K log P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

? equation 1

At time t = 100 s, P_t = 0.6 atm and P₀ = 0.5 atm,

Substituting in equation 1,

100 = 2.303/k log 0.5 / (2X0.5) - 0.6

Thus, k = 2.231 * 10^-3 s^-1

The rate of reaction R = k * P_S02Cl₂

When total pressure P_t = 0.65 atm and P₀ = 0.5 atm, then

[P_S02Cl₂ = 2P₀-P_t

Thus, substituting the values, P_S02Cl₂ = 2 (0.5)-0.6 = 0.35 atm

R = k * P_S02Cl₂ = 2.231 * 10^-3 s^–1 * 0.35

Rate of the reaction R = 7.8 * 10^-4atm s^–1

4.29 When t = 0, the total partial pressure is P₀ = 35.0 mm of Hg

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀ Hence, P₀-p = P_t-2 ( P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time 't'

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

t= 2.303/K log P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

? equation 1

At time t = 360 s, P_t = 54 mm of Hg and P₀ = 30 mm of Hg, Substituting in equation 1,

360 = 2.303/k log 30 / (2X30)-54

k= 2.175 X 10^-3 s^-1

At time t = 720 s, P_t = 63 mm of Hg and P₀ = 30 mm of Hg, Substituting in equation 1,

720 = 2.303 / k log 30 / (2X30) - 63

Thus, k = 2.235 * 10^-3 s^–1

Taking average, k = (2.235 * 10^-3 s^–1 + 2.175 * 10^-3 s^–1)/2

? k = 2.21 * 10^-3 s^–1.

7.70

As Bond dissociation energy generally decreases on moving down the group as the atomic size of the element increases. However, among halogens, the bond dissociation energy of F₂ is lower than that of Cl₂ and F₂ due to the small atomic size of

Thus increasing order for bond dissociation energy among halogens is as follows:I₂2

As Bond dissociation energy of H-X molecules where X is the halogen decreases with increase in the atomic HI is the strongest acid as it loses H atom easily due to weak bonding between H and I.

So Increasing acid strength is as follows: HF

Basic strength decreases as we move from Nitrogen to Bismuth down the group as the size of the atom increases which is electron density of the atom decreases.

So Basic Strength is as follows: BiH3