Which expression is the integrated rate law for a zero-order reaction?

Which expression is the integrated rate law for a zero-order reaction?

• A. [A]t = -kt + [A0]
• B. 1/[A]t = kt + 1/[A0]
• C. [A]t = [A0] / (1 + k t)
• D. ln[A]t = -kt + ln[A0]

Answer: (A)

For a zero-order reaction, the rate is constant and does not depend on how much A is present. This means -d[A]/dt = k, with k being a constant. Integrating gives [A]t = [A]0 - kt. So the concentration decreases linearly with time, and the integrated rate law is [A]t = -kt + [A0]. This is the best expression because it directly reflects the constant-rate nature of zero-order kinetics. The other forms correspond to different orders: for first-order, ln[A]t = -kt + ln[A0]; for second-order, 1/[A]t = kt + 1/[A0]. The form [A]t = [A0]/(1 + kt) does not match the standard integrated law for zero-order (and would require a different rate expression or mis-specified constants).

For a zero-order reaction, the rate is constant and does not depend on how much A is present. This means -d[A]/dt = k, with k being a constant. Integrating gives [A]t = [A]0 - kt. So the concentration decreases linearly with time, and the integrated rate law is [A]t = -kt + [A0].

This is the best expression because it directly reflects the constant-rate nature of zero-order kinetics. The other forms correspond to different orders: for first-order, ln[A]t = -kt + ln[A0]; for second-order, 1/[A]t = kt + 1/[A0]. The form [A]t = [A0]/(1 + kt) does not match the standard integrated law for zero-order (and would require a different rate expression or mis-specified constants).