chemistry question (kinetics): please Help. I got a final in 2 days :( 1. In a s

Question

chemistry question (kinetics): please Help. I got a final in 2 days :(

1. In a second order reaction the half-life depends on

a) initial concentration b)temperature c)both

2. which of the following parameters depends on temperature?

a) rate constant b)equilibrium c) both

3. a catalyst will affect which of the following parameters?

a)equilibrium constant b)activation energy c) none of these

4. CO(g) + 3H2 (g) -> CH4(g) +H2O (g)

you have 1000ml flask which has 0.891 mol of CO and 0.567 mol of H. After 7.6 secs, 0.123 mol of H2O are formed. what concentration of H was consumed in the first 7.6 secs?

5. A reaction has a rate constant (k) of 3.0x10^-4 s^-1 at 298K and activation energy of 100 kJ/mol.

a) what is the value of the rate constant (in s^-1) at 35 degree C.

b) whats the value of frequency factor (in s^-1)?

c)if the addition of catalyst lowered the activation energy by 10kJ/mol, at what temperature (in K) could the catalyzed reaction be run to realize the same rate as the un-catalyzed reaction run at 298K?

5.

O2(g)ß -> 2O(g)                                               (k1,k-1)

O(g)+N2(g) -> NO(g) +N (g)                                (k2)

N(g) +O(g) -> NO(g)                                           (k3)

based on the graph and information above, how to determine the slow step? (which one is this step?)

write the overall reaction given the above reaction mechanism. Identify any intermediate and/or catalytic species.

Determine the rate law from the reaction mechanism above. Use the specific rate constants (eg: K1, K-1, K2, K3, do not combine them)

Draw a graph to show how a catalyst would change the reaction.

Explanation / Answer

Solutions:

1. For a second order reaction,

half life t1/2 can be written as = 1/k[A]o, where,

k = rate constant which is directly dependent on temperature

[A]o = initial concentration

So, the half life for a second order reaction is dependent on,

c) both

2. Both rate constant and equilibrium are dependent on temperature. Thus the correct answer would be,

c) both

For rate constant : An increase in temperature will always increase the rate of a reaction since the particles will have more kinetic energy and there would be more productive collisions. The value of the rate constant will hence increase. The increase in rate constant with temperature is exponential, meaning that every time the temperature increases by a certain number of degrees, the rate constant increases by a fixed factor.

For equilibrium : The effect of temperature on equilibrium can be product favoured or reactant favoured depending on the type of reaction. For an endothermic reaction, the increase in temperatur would increase rate and hence the equilibrium is achived faster. For exothermic reaction an increase in temperature would favor reactants. This is based of LeChatellier's principle.

3. A catalysts will effect,

b)activation energy

The catalysts binds to the substrate and thus the energy required for the reactant to form products reduce by decreasing the activation energy.

4. Given here is,

CO(g) + 3H2 (g) -> CH4(g) +H2O (g)

We have 1000ml flask which has initially at time t = 0,

0.891 mol of CO and 0.567 mol of H.

at t = 7.6 secs, 0.123 mol of H2O are formed

Now looking at the above reaction equation, 1 mole of CO reacts with 3 mols of H2 to form 1 mole of H2O

Thus, when 0.123 mols of H2O if formed, the concentration of H2 consumed would be,

= 0.123 x 3 = 0.369 mols of H2 is consumed after 7.6 secs.

5. Given here reaction has a rate constant (k) of 3.0x10^-4 s^-1 at 298K and activation energy of 100 kJ/mol.

a) To calculate a) the value of the rate constant (in s^-1) at 35 degree C.

Using the relation,

ln(k2/k1) = Ea/R [1/T1-1/T2]

where,

k1 = 3.0 x 10^-4 s^-1

k2 = unknown

T1 = 298 K

T2 = 35 oC = 308 K

Ea = 100 kJ/mol

R = gas constant

Feed the values,

ln(k2/3.0 x 10^-4) = 100000/8.314[1/298-1/308]

solving for k2,

k2 = 1.11 x 10^-3 s^-1 is rate constant at 35 oC

b) To calculate frequency factor A

relation to be used,

lnk = lnA - Ea/RT

taking k = 3.0 x 10^-4 s^-1

Ea = 100 kJ/mol

R = gas constant

T = 298 K

Feed values,

ln(3.0 x 10^-4) = lnA - 100000/8.314 x 298

solving for A,

A = 1.014 x 10^14 s^-1 is the value of frequency factor

c) When the activation energy was reduced by 10 kJ/mol, we have,

Ea = 90 kJ/mol

R = gas constant

T = unknown

k = 3.0 x 10^-4 s^-1

Feed in the equation,

lnk = lnA - Ea/RT

ln(3.0 x 10^-4) = ln(1.014 x 10^14) - 90000/8.314T

solving for T,

T = 268.202 K temperature the rate would be same as at 298 K

6. Looking at the above reaction and graph,

we can say that the third step N(g) + O(g) --> NO(g) is the slow step.

Combining the equations, we get overall reaction as,

N2(g) + O2(g) ----> 2NO(g)

The rate law thus can be definied in terms of the reaction as,

rate = k3[N][O] or k3[N2][O2] to be more specific

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