A)if it takes 0.20 min to decompose 15% of a 0.300 m solution of nitrosyl chloride, what is k(the rate constant)? the decomposition of nitrosyl chloride is a second order reaction: nocl(> no(g)+1/2cl2(g)b)calculate the activation energy? when the temperature increases from 15c to 25c for a certain reaction, its rate constant doubles. calculate the activation energy, ea+ (remember: with r=8.31 j/mol. k, ea must be expressed in j/mol)

a) k = 0.0417 M⁻¹.s⁻¹

b) Ea = 49.4 kJ/mol

Explanation:

a) If the decomposition reaction is a second-order reaction, the rate law is:

rate = k*[NOCl]²

The rate is how much the concentration decays per second (M/s), thus, if its initially 0.300 M and it decays 15%, the final concentrations 85% of 0.300, which is 0.255 M. The time needed for that is 0.20 min = 12 s, so:

rate = (0.300 - 0.255)/12

rate = 0.00375 M/s

Thus, the rate constant is:

0.00375 = k*(0.300)²

0.09k = 0.00375

k = 0.0417 M⁻¹.s⁻¹

b) By the Arrhenius equation, the constant k is related to the activation energy as:

k =

Where A is a constant, R is the gas constant (8.31 J/mol.K) and T is the temperature. If we apply ln in the equation:

ln(k) = ln(A) - Ea/RT

Thus, for T1 = 15°C (288 K), k = k1, and for T = 25°C (298 K), k = k2, knowing that Ea and A are independent of the temperature:

ln(k1) = ln(A) - Ea/RT1

ln(k2) = ln(A) - Ea/RT2

ln(k1) - ln(k2) = ln(A) - ln(A) - Ea/RT1 + Ea/RT2

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

If the rate constant doubles, k2 = 2k1

ln (1/2) = (Ea/8.31) * (1/298 - 1/288)

-0.6932 = (Ea/8.31)*(-1.1652x10⁻⁴)

1.402x10⁻⁵Ea = 0.6932

Ea = 49438.84 J/mol

Ea = 49.4 kJ/mol

the correct answer to this question is 2