The rate constant of the elementary reaction CH3OCH3(g) CH4(g) +CH2O(g) is k = 8.33×10-6 s-1 at 427°C, and the reaction has an activation energy of 245 kJ mol-1. (a) Compute the rate constant of the reaction at a temperature of 545°C. s-1 (b) At a temperature of 427°C, 8.32×104 s is required for half of the CH3OCH3 originally present to be consumed. How long will it take to consume half of the reactant if an identical experiment is performed at 545°C?

(a) The rate constant is 3.61×10^-3 s^-1

(b) 7.12×10^4 s

Explanation:

(a) Log (K2/K1) = Ea/2.303R × [1/T1 - 1/T2]

K1 = 8.33×10^-6 s^-1

Ea = 245 kJ = 245,000 J

R = 8.314 J/mol.K

T1 = 427°C = 427+273 = 700 K

T2 = 545°C = 546+273 = 818 K

Log (K2/8.33×10^-6) = 245,000/2.303 × [1/700 - 1/818]

Log (K2/8.33×10^-6) = 2.637

K2/8.33×10^-6 = 10^2.637

K2 = 8.33×10^-6 × 433.51 = 3.61×10^-3 s^-1

(b) The relationship between temperature and the time required for reactants to be consumed is inverse

t2 = T1t1/T2

T1 = 427 °C = 700 K

t1 = 8.32×10^4 s

T2 = 545 °C = 818 K

t2 = 700×8.32×10^4/818 = 7.12×10^4 s

The enthalpy of the reaction in kJ/mol is 39.7 kJ/mol.

The equation of the reaction is;

HCl(aq) + KOH(aq) > KCl(aq) + H2O(l)

We have the following information from the question;

Final volume of solution = 100.00 ml HCl + 100.00 ml KOH = 200.00 ml

Mass of solution = 200.00 g of solution

Number of moles of solution = 0.500 M × 100.00/1000 L = 0.05 moles

Notice that the reaction is 1:1

Final temperature of solution = 25.5°C

Initial temperature of the acid and base = 23.0°C

Specific heat of the solution = 3.97 J. g-1 °C

Using the formula;

ΔH = mcθ

ΔH = Heat of reaction

m = mass of solution

c = Specific heat of the solution

θ = temperature rise

Substituting values;

ΔH = 200.00 g × 3.97 J. g-1 °C ×  (25.5°C -  23.0°C)

ΔH = 1.985 KJ

Enthalpy of the reaction in kJ/mol =  1.985 KJ/0.05 moles

= 39.7 kJ/mol

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