The activation energy, ea, for the reaction 2 n2o5 (g) latex: \longrightarrow ⟶ 4 no2 (g) + o2 (g) is 22 kj/mol. what is the rate constant at 84.8°c if the rate constant is 1.868 sec-1 at 16.6°c? enter to 3 decimal places. latex: \ln\frac{k2}{k1}=\frac{ea}{r}\left( \frac{1}{t1}\: -\frac{1}{t2}\right)

10.37 s-1

Explanation:

From

k= A e-^Ea/RT

Given

Ea=22KJmol-1

T=16.6+273= 289.6K

k= 1.868 sec-1

R= 8.314JK-1mol-1

A???

Hence

A= k/e^-Ea/RT

A= 1.868/e-(22000/8.314×289.6)

A= 1.7 ×10^4

Substitute into to find k at 84.8°C

k= 1.7×10^4× e-(22000/8.314×357.8)

k=10.37 s-1

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

m=9

step-by-step explanation: