Need for Speed
What are the units of the rate constant k for a zero-order reaction?
What is M/s (or mol/(L*s))?
Rate always has units of M/s.
Rate = k[A]^0. Since [A]^0 = 1,
Rate = k(1). Therefore, k must have the exact same units as Rate: M/s
What is the defining characteristic of a first-order half-life compared to zero or second-order?
What is constant and independent of the initial concentration?
formula t_1/2 = ln(2)/k = 0.693/k.
There is no [A]_0 term in the equation, meaning concentration doesn't affect the time it takes to halve.
Define "molecularity" and give the specific term for an elementary step involving exactly one reactant molecule.
Molecularity is the number of reactant molecules participating in an elementary step; Unimolecular.
At chemical equilibrium, what is dynamically true about the forward and reverse reactions?
The rates of the forward and reverse reactions are equal.
*Remember that concentrations do not have to be equal.*
If a forward reaction has an equilibrium constant of K = 10, what is the K for the reverse reaction?
What is 0.1?
Reversing an equation means you take the inverse of K: K_reverse = 1 / K_forward. 1 / 10 = 0.1.
For the reaction 2A -> 3B, if the rate of decomposition of A is 0.40 M/s, what is the rate of formation of B?
What is 0.60 M/s?
Rate = -1/2(delta[A]/delta t) = 1/3(delta[B]/delta t). The decomposition of A is 0.40. Therefore, 1/3(delta[B]/delta t) = 1/2(0.40) = 0.20. Multiply both sides by 3 to get delta[B]/delta t = 0.60 M/s.
A first-order reaction has a rate constant k = 0.0346 s^-1. What is its half-life?
What is 20.0 s?
t_1/2 = 0.693 / k = 0.693 / 0.0346 = 20.0 s.
In a multi-step reaction mechanism, what is the definition of an intermediate?
A chemical species that is produced in one elementary step and completely consumed in a subsequent step.
*Reminder: This is the opposite of a catalyst.*
What are the units for the equilibrium constant, Kc?
What is there are no units?
Equilibrium constants are derived from activities, the ratio of a concentration to its standard state, meaning units cancel out completely.
If a balanced chemical equation is multiplied by a coefficient of 2, what mathematical operation must you perform on its equilibrium constant K?
What is you must square it (K^2)?
Which plot yields a straight line for a second-order reaction, and what is its slope?
What is 1/[A] vs time?; the slope is positive k
The integrated rate law is 1/[A]_t = kt + 1/[A]_0. This matches y = mx + b, where y is 1/[A], x is t, and the slope m is positive k.
If a first-order reaction takes 40 minutes to drop from 1.0 M to 0.25 M, what is its half-life?
What is 20 minutes?
1.0 M -> 0.50 M (1 half-life). 0.50 M -> 0.25 M (2nd half-life). Total of 2 half-lives = 40 minutes. 40 / 2 = 20 minutes.
According to the Arrhenius equation, if the activation energy (Ea) of a reaction increases, what happens to the rate constant k?
What is k decreases? (The reaction gets slower).
k = A * e^(-Ea/RT). Because Ea is negative in the exponent, making it larger makes the exponent a larger negative number. A larger negative exponent makes the whole term smaller.
If K = 4.5 x 10^-6, does the reaction vessel contain mostly reactants or mostly products at equilibrium?
What is mostly reactants?
K = [Products]/[Reactants]. A very small K (K << 1) means the denominator (reactants) is vastly larger than the numerator (products).
For a reaction, the forward rate constant k_f = 0.040 and the reverse rate constant k_r = 0.010. What is the equilibrium constant K_eq?
What is 4.0?
At equilibrium, Rate_f = Rate_r, meaning k_f[Reactants] = k_r[Products].
Rearranging gives [Products]/[Reactants] = k_f / k_r.
Therefore, K = 0.040 / 0.010 = 4.0.
Given this initial rates data: Exp 1 ([A] = 0.10 M, Rate = 0.020 M/s) and Exp 2 ([A] = 0.20 M, Rate = 0.080 M/s). What is the reaction order with respect to A?
What is Second order?
Set up the ratio: Rate_2 / Rate_1 = ([A]_2 / [A]_1)^x. Plug in the numbers: 0.080 / 0.020 = (0.20 / 0.10)^x. This simplifies to 4 = 2^x. Therefore, x = 2.
A chemical decomposes via first-order kinetics with k = 0.10 days^-1. If you start with 500 mg, how much remains after 14 days?
What is 123 mg?
Use ln[A]_t = -kt + ln[A]_0. Plug in values: ln[A]_t = -(0.10)(14) + ln(500). ln[A]_t = -1.4 + 6.214 = 4.814. To get rid of the natural log, use e: [A]_t = e^4.814 = 123.2 mg.
A reaction has k = 1.2 x 10^-2 s^-1 at 300 K and k = 3.5 x 10^-2 s^-1 at 310 K. Set up the exact equation needed to solve for Ea.
ln(3.5 x 10^-2 / 1.2 x 10^-2) = (Ea / 8.314) * (1/300 - 1/310).
The formula: ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2). Remember to use R = 8.314 J/(mol*K), NOT the gas-law R.
What two physical states are always excluded from the equilibrium constant expression?
Pure solids (s) and pure liquids (l).
For the Haber process N2(g) + 3H2(g) <=> 2NH3(g), calculate Kc if equilibrium concentrations are [N2] = 1.0 M, [H2] = 2.0 M, and [NH3] = 4.0 M.
What is 2.0?
Write the expression first: Kc = [NH3]^2 / ([N2][H2]^3).
Plug in the numbers: Kc = (4.0)^2 / ((1.0)(2.0)^3) = 16 / ((1.0)(8)) = 2.0.
If a reaction is first-order in A and second-order in B, what happens to the overall reaction rate if [A] is doubled and [B] is halved?
The rate is halved (reduced by a factor of 2).
Rate_1 = k[A]^1[B]^2. Plug in the changes for Rate_2 = k(2[A])(1/2[B])^2. Square the one-half: k(2[A])(1/4[B]^2). Multiply the coefficients: 2 * 1/4 = 1/2. The new rate is half the original.
What is the rate constant k if it takes 55 seconds for a first-order reaction to be 75% complete?
What is 0.025 s^-1?
If it is 75% complete, 25% remains. So, [A]_t = 0.25[A]_0. Use the ratio form: ln([A]_t / [A]_0) = -kt. Plug in the ratio: ln(0.25) = -k(55). -1.386 = -55k. k = 0.025 s^-1.
Solve the previous equation for Ea in kJ/mol.
What is ~ 83 kJ/mol?
1. ln(2.916) = (Ea/8.314)*(0.003333 - 0.003225)
2. 1.07 = (Ea/8.314)*(0.000107)
3. Multiply 8.314 over: 8.896 = Ea(0.000107)
4. Divide: Ea = 83,140 J/mol. Convert to kJ: 83.1 kJ/mol.
Write the correct Kc expression for the following reaction: CaCO3(s) <=> CaO(s) + CO2(g)
Kc = [CO2]
Cross out CaCO3 because it is solid. Cross out CaO because it is solid. The only thing left is CO2 gas. Since its coefficient is 1, Kc = [CO2]^1.
Reaction 1 has K1 = 2. Reaction 2 has K2 = 3. If you add Reaction 1 and Reaction 2 together to form a new overall reaction, what is the K3 of this consecutive equilibrium?
What is 6?
When adding individual chemical steps together to get an overall reaction, you must multiply their respective equilibrium constants. K3 = K1 * K2 = 2 * 3 = 6.