Determine the rate constant for each of the following first-order reactions. In each case, write the rate law for the rate of loss of A.

(a) 2 A B + C, given that the concentration of A decreases to one-fourth its initial value in 57 min

(b) 2 A B + C, given that A0 = 0.036 mol·L-1 and that after 85 s the concentration of B increases to 0.0095 mol·L-1

(c) 2 A 3B + C, given that A0 = 0.068 mol·L-1 and that after 8.2 min the concentration of B increases to 0.030 mol·L-1

See your earlier post.

a) 2A --> B + C

b) 2A --> B + C

c) 2A --> 3B + C

To determine the rate constant for each of the first-order reactions and write the rate law for the rate of loss of A, we need to use the integrated rate law for a first-order reaction.

The integrated rate law for a first-order reaction is given by:

ln(A/A0) = -kt

Where A is the concentration of A at a given time, A0 is the initial concentration of A, k is the rate constant, and t is the time.

(a) For the reaction 2A -> B + C, given that the concentration of A decreases to one-fourth its initial value in 57 min, we can plug in the values into the integrated rate law equation:

ln(A/0.25A0) = -k(57 min)

Simplifying the equation, we have:

ln(1/0.25) = -k(57 min)

ln(4) = -k(57 min)

Now, we can solve for the rate constant k:

k = -ln(4)/57 min

(b) For the reaction 2A -> B + C, given that A0 = 0.036 mol·L-1 and that after 85 s the concentration of B increases to 0.0095 mol·L-1, we can plug in the values into the integrated rate law equation:

ln(A/0.036) = -k(85 s)

Simplifying the equation, we have:

ln(B/0.036) = -k(85 s)

ln(0.0095/0.036) = -k(85 s)

Now, we can solve for the rate constant k:

k = -ln(0.0095/0.036)/85 s

(c) For the reaction 2A -> 3B + C, given that A0 = 0.068 mol·L-1 and that after 8.2 min the concentration of B increases to 0.030 mol·L-1, we can plug in the values into the integrated rate law equation:

ln(A/0.068) = -k(8.2 min)

Simplifying the equation, we have:

ln(B/0.068) = -k(8.2 min)

ln(0.030/0.068) = -k(8.2 min)

Now, we can solve for the rate constant k:

k = -ln(0.030/0.068)/8.2 min

The rate law for the rate of loss of A in all three reactions would be expressed as:

Rate = -k[A]