We read your photo as:

A) Order of reaction: The order of reaction with respect to a reactant is the exponent to which its concentration term is raised in the rate law. It indicates how the rate of reaction is affecte

Chemistry

ScanSolve AI Answer

Answers

2 parts
e(ii)

8×103 s8 \times 10^3 \text{ s}

e(iii)

first order

Here are the solutions to the questions:

e)

i)
A) Order of reaction: The order of reaction with respect to a reactant is the exponent to which its concentration term is raised in the rate law. It indicates how the rate of reaction is affected by a change in the concentration of that specific reactant. The overall order of reaction is the sum of the orders with respect to all reactants.

B) Instantaneous reaction rate: This is the rate of a chemical reaction at a particular moment in time. It is determined by calculating the slope of the tangent to the concentration-time curve at that specific time.

ii) Determine the half-life of the reaction:
The half-life (t1/2t_{1/2}) is the time required for the concentration of a reactant to decrease to half of its initial value.
Let's examine the data:

  • From [A]=0.40 mol dm−3[A] = 0.40 \text{ mol dm}^{-3} (at t=8×103 st = 8 \times 10^3 \text{ s}) to [A]=0.20 mol dm−3[A] = 0.20 \text{ mol dm}^{-3} (at t=16×103 st = 16 \times 10^3 \text{ s}). The concentration has halved. The time taken is (16−8)×103 s=8×103 s(16 - 8) \times 10^3 \text{ s} = 8 \times 10^3 \text{ s}.
  • From [A]=0.28 mol dm−3[A] = 0.28 \text{ mol dm}^{-3} (at t=12×103 st = 12 \times 10^3 \text{ s}) to [A]=0.14 mol dm−3[A] = 0.14 \text{ mol dm}^{-3} (at t=20×103 st = 20 \times 10^3 \text{ s}). The concentration has halved. The time taken is (20−12)×103 s=8×103 s(20 - 12) \times 10^3 \text{ s} = 8 \times 10^3 \text{ s}.
  • From [A]=0.58 mol dm−3[A] = 0.58 \text{ mol dm}^{-3} (at t=4×103 st = 4 \times 10^3 \text{ s}), half of this is 0.29 mol dm−30.29 \text{ mol dm}^{-3}. This value is very close to 0.28 mol dm−30.28 \text{ mol dm}^{-3} at t=12×103 st = 12 \times 10^3 \text{ s}. The time taken is approximately (12−4)×103 s=8×103 s(12 - 4) \times 10^3 \text{ s} = 8 \times 10^3 \text{ s}.

Since the half-life is constant, the reaction is first order.
The half-life of the reaction is 8×103 s\boxed{8 \times 10^3 \text{ s}}.

iii) Determine the order of the reaction with respect to A:
As observed in part (ii), the half-life of the reaction is constant and independent of the initial concentration of A. This is a characteristic property of a first-order reaction.
Therefore, the order of the reaction with respect to A is firstorder\boxed{first order}.

d) Define an acid according to Brønsted and Lowry:
According to the Brønsted-Lowry theory, an acid is a proton donor. It is a chemical species that donates a hydrogen ion (H+\text{H}^+) to another substance in a chemical reaction.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp