Question

When the chemical reaction 2no(g)+o2(g)→2no2(g) is carried out under certain conditions, the rate of disappearance of no(g) is 5. 0×10−5 ms−1. What is the rate of disappearance of o2(g) under the same conditions?.

Answer 1

The rate of disappearance of O2(g) under the same conditions is 2.5 × 10⁻⁵ m s⁻¹.

What is the rate law of a chemical equation?

The rate law of a chemical reaction equation is usually dependent on the concentration of the reactant species in the equation.

The chemical reaction given is;

$\mathbf{2 NO_{(g)} + O_{2(g)} \to 2 NO_{2(g)} }$

The rate law for this reaction can be expressed as:

$\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}= +\dfrac{1}{2}\dfrac{d[NO_2]}{dt}}$

Recall that:

• The rate of disappearance of NO(g) = 5.0× 10⁻⁵ m s⁻¹.

• Since both NO and O2 are the reacting species;

Then:

• The rate of  disappearance of NO(g) is equal to the rate of  disappearance of O2(g)

$\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}}$

$\mathbf{= -\dfrac{1}{2} \times 5.0 \times 10^{-5} = rate \ of \ disappearance \ of \ O_2}$

Thus;

The rate of disappearance of O2 = 2.5 × 10⁻⁵ m s⁻¹.

Therefore, we can conclude that two molecules of NO are consumed per one molecule of O2.

Learn more about the rate law here:

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