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k0ka [10]
3 years ago
9

A chemical reaction involves reactant species A, B, and C. Leaving all other factors identical, doubling the concentration of sp

ecies B increases the rate by a factor of 4. The rate of reaction is not affected by changing the concentration of species A. The rate of the reaction is linearly dependent on the concentration of C. What is the rate equation for this reaction?Rate=k[A][B]2[C]Rate=k[A]2[B][C]2Rate=k[A][B][C]Rate=k[B]2[C]Rate=k[B]2[C]2None of the above
Chemistry
1 answer:
umka21 [38]3 years ago
3 0

Answer:

Rate = k . [B]² . [C]

Explanation:

The dependence of the reaction rate on the concentration of the reactants is given by the reaction order of each one, as shown in the rate equation.

Rate=k.[A]^{x} .[B]^{y} .[C]^{z}

where,

k is the rate constant

x, y, z are the reaction orders.

  • <em>The rate of reaction is not affected by changing the concentration of species A.</em> This means that the reaction order for A is x = 0 since when its concentration changes, the rate stays the same.
  • <em>Leaving all other factors identical, doubling the concentration of species B increases the rate by a factor of 4.</em> This means that the reaction order for B is y = 2, so when the concentration is doubled, the new rate is 2² = 4 times the initial rate.
  • The rate of the reaction is linearly dependent on the concentration of C. This means that the reaction order for C is z = 1, that is, a linear dependence.

All in all, the rate equation is:

Rate = k . [B]² . [C]

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The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
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Answer:

3,964 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

7 0
4 years ago
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