Question

CHEM1034 Ex 3A-written part 1. Given the activation energy E, of 36 88 kcal/imol, what's the percentage inscreas of rate cosan k by raising the reaction temperature from 75 "C to 100 2. Without detailed calculations, predict and explain in detail the sign of AS for the system for 3. When magnesium sulfite decomposes,the solid transforms into magnesium oxide and sulfur dioxide. At what temperature will his reaction be spontaneous according to Gibb's Energy? dH in kl/mol for: MgSOJs)--1068, MgO(s)--601.8, sois)--2968 AS in J/mol K for: MgSOs(s)- 121, Mgo(s)- 26.9, SO(g)-248.1 s. How many ways you can make a reaction go faster? Explain in details why such wayís) can speed up the reaction TOwae erecction ap tister For a 2ndorder reaction, how can you plot of concentration with respect to time to producesa straight line? Justify your answer. 5.

Answer 1

1)

According to The Arrhenius equation

$k=Ae^{-\frac{E_{a}}{RT}}$

initial temparature (T₁) = 75⁰C = 273+75 K = 348 K

final temparature (T₂) = 100⁰C = 273 + 100 K = 373 K

Activation energy (E_a) = 36.88 kcal/mol

universal gas constant (R) = 2 cal K^-1 mol^-1

percentage change in k = $\frac{k_{2}-k_{1}}{k_{1}}\times 100$

=  $\frac{Ae^{\frac{-E_{a}}{RT_{2}}}-Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{1}}}}\times 100$

=  $\frac{A(e^{\frac{-E_{a}}{RT_{2}}}-e^{\frac{-E_{a}}{RT_{1}}})}{Ae^{\frac{-E_{a}}{RT_{1}}}}\times 100$

= $\frac{e^{\frac{-E_{a}}{RT_{2}}}-e^{\frac{-E_{a}}{RT_{1}}}}{e^{\frac{-E_{a}}{RT_{1}}}}\times 100$

=  $\frac{e^{\frac{-36.88}{2\times 373}}-e^{\frac{-36.88}{2\times 348}}}{e^{\frac{-36.88}{2\times 348}}}\times 100$

=  $\frac{e^{-0.049}-e^{-0.052}}{e^{-0.052}}\times 100$

=  $\frac{0.952-0.949}{0.949}\times 100$

= 0.0031 $\times$ 100

= 3.1 %

2) For the reaction Mg(s) + O₂(g) $\rightarrow$ MgO(s)

one of the reactants oxygen is in gaseous state while the products are in solid state.Entropy is measure of disorderness of a system.since gaseous reactant is converted to solid product disorderness decreases as a result sign of  $\Delta$S will be negative

3) given reaction MgSO₃(s) $\rightarrow$ MgO(s) + SO₂ (g)

for a reaction to be spontaneous $\Delta$G 0

$\Delta$H of the reaction =  $\Delta$H of MgO +$\Delta$​​​​​​​H of SO₂ -$\Delta$​​​​​​​H of MgSO₃

  = -601.8 +(-296.8) - (-1068)

= 169.4 kJ/mol

$\Delta$S of the reaction = $\Delta$S of MgO +$\Delta$S of SO₂ -$\Delta$S of MgSO₃

= 26.9 + 248.1 - 121

= 154 J/mol

= 0.154 kJ/mol

$\Delta$G = $\Delta$H-T$\Delta$S

for a reaction to be spontaneous $\Delta$G 0

$\Delta$H-T$\Delta$S 0

169.4 - 0.154T 0

  T $\frac{169.4}{0.154}$

T 1100 K

therefore over 1100 K the reaction is spontaneous

4)There are four ways of increasing the rate of a reaction

1.Increase the temperature - this increases the kinetic energy of particles, so when they collide with each other the collision has a greater energy so is more likely to overcome the activation energy barrier to a reaction. This means a greater proportion of collisions will be successful and so the rate of reaction will be faster.

2.Increase the concentration of reactants - this will Increase the number of molecules in the same space, therefore more collisions will occur between molecules so the rate of reaction will be faster. This is also true for increasing the pressure of a gas.

3.Increase the surface area - if the reaction involves a solid, it can be broken down into smaller parts to increase its surface area. This means more molecules will be available to react and so the reaction will be faster.

4.Use a catalyst - a catalyst is a substance which increases the rate of a reaction without being used up in the reaction. It does this by providing an alternative pathway for the reaction which has a lower activation energy and therefore takes place more quickly.

5)

The rate law of a second order reaction is

$\frac{d[A]}{dt} = -k[A]^{2}$

this can be rearranged as

$\frac{d[A]}{[A]^{2}} = -kdt$

integrating on both sides

with [A_t] as concentration of A at time t and [A₀] as initial concentration of A

we get

$\frac{1}{[A_{t}]} = \frac{1}{[A_{0}]} + kt$

this will be a linear plot between $\frac{1}{[A_{t}]}$ and