How Does Temperature Affect Reaction Rates and Molecular Interactions?

[courtrigrad]

1.How would temperature effect the rate of a reaction? Would it increase its kinetic energy thereby speeding up the reaction. Is it right in saying that as the temperature increases, the frequency of collisions increases between molecules, and the orientation changes?

2. If we have 4Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3} and we know that the concentration of O_{2} changes 0.05 mol/L every 2 seconds what is the rate of disappearance of Fe in mol/L*sec? So -\frac{1}{4} \frac{\Delta[Fe]}{\Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = \frac{1}{2} \frac{\Delta[Fe_{2}O_{3}]}{\Delta t}. In one second the rate of disappearance of O_[2} would be 0.025 mol/L. So would it be \frac{0.025}{3} for the rate of disappearance of Fe? For the rate of evolution of Fe_{2}O_{3} would be 2 \times 0.025?

3. If we have a reaction between X and Z and the rate law is k[X]^{2}[Z]^{3} what would happen if:
a. concentration of X doubled while Z remains constant
b. concentration of Z triples while X remains constant
c. concentrations of both X and Z are doubled

For (a) would the rate quadruple because X has a reaction order of 2? For (b) would the rate go up by a factor of 8? For (c) would the rate go up by a factor of 16?

 

[bross7]

1. The orientation does not necessarily have to change, but you are correct in assuming the frequency of collisions increases. As the temperature increases the kinetic energy of the molecules increases which = more collisions. The increased number of collisions increases the likelihood that a collision will occur at the proper orientation with the proper activation energy.

2. Not quite. Keep in mind that the rates being made equivalent are coming from their ratios in the initial equation.

Writing the formula like this: -\frac{1}{3}\frac{\Delta[O_2]}{\Delta T}=\frac{1}{2}\frac{\Delta [Fe_2O_3]}{\Delta T}

Is very similar to writing:

\frac{2}{3}\Delta[O_2]=\Delta [Fe_2O_3]

While not specifically the same as the equation when solving for the mole ratio, I am just trying to draw on the comparison.
When comparing the rates, you need to take into account the ratio's of each reactant/product when finding the comparable rates.

The rate of disappearance of O_2 is just the value for \Delta [O_2]. When finding the rates for the other two substances you need to factor in the numerical ratios aswell.

3. a) Yes it would quadruple k(2X)^2 (Z)^3 = 4k(X^2)(Z^3)
b) Careful, it is the Z concentration that is tripled (3Z)^3 = ...
c) k(2X)^2 (2Z)^3 = (4)(8)k(X^2)(Z^3)

 

[thepatientmental]

1. Temperature can definitely affect the rate of a reaction. As temperature increases, the kinetic energy of the molecules also increases, leading to more frequent and energetic collisions between molecules. This results in a higher chance of successful collisions and therefore, a faster reaction rate. Additionally, as the temperature increases, the orientation of the molecules may also change, making it easier for them to collide in a way that leads to a reaction. So, it is correct to say that an increase in temperature can increase the frequency of collisions and change the orientation of the molecules, ultimately speeding up the reaction.

2. Yes, you are correct in your calculations. The rate of disappearance of Fe would be 0.025 mol/L*sec and the rate of evolution of Fe_{2}O_{3} would be 2*0.025 mol/L*sec. This is because the stoichiometric coefficients in the balanced equation determine the ratio at which the concentrations of the reactants and products change.

3. For (a), the rate would actually increase by a factor of 4, not quadruple. This is because the rate law is dependent on the concentration of X squared, so doubling the concentration would result in a rate increase of 2 squared, which is 4. Similarly, for (b), the rate would increase by a factor of 27, not 8. And for (c), the rate would increase by a factor of 64, not 16. This is because the rate law is dependent on both the concentrations of X and Z, and doubling both concentrations would result in a rate increase of 2 squared times 2 cubed, which is 64.