For the reaction given below, the frequency factor A is 8.7 1012 s−1 and the activation energy is 63 kJ/mol. NO(g) + O3(g) → NO2(g) + O2(g) What is the rate constant for the reaction at 70°C?

K(70°C) = 2213.376 s-1

Explanation:

balanced reaction:

∴ Ea = 63 KJ/mol

∴ A = 8.7 E12 s-1

∴ T = 70°C ≅ 343 K

Arrhenius eq:

∴ R = 8.314 E-3 KJ/K.mol

⇒ K(70°C) = (8.7 E12 s-1)e∧[-(63)/(8.314 E-3)(343)]

⇒ K(70°C) = (8.7 E12 s-1)*(2.5444 E-10)

⇒ K(70°C) = 2213.376 s-1

the answer is c i believe

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2