Question:
A → P is a first order reaction. At T(K), the concentration of reactant (A) after 10 min of the reaction is \(x\) mol L\(^{-1}\). After 20 min, the concentration of A is \(y\) mol L\(^{-1}\). What is its rate constant (in min\(^{-1}\))?
A → P is a first order reaction. At T(K), the concentration of reactant (A) after 10 min of the reaction is \(x\) mol L\(^{-1}\). After 20 min, the concentration of A is \(y\) mol L\(^{-1}\). What is its rate constant (in min\(^{-1}\))?
Show Hint
Rate constant for first order reaction can be calculated using concentration changes over time.
Updated On: Jun 2, 2025
- \(0.2303 \log \frac{x}{y}\)
- \(2.303 \log \frac{x}{y}\)
- \(2.303 \log \frac{y}{x}\)
- \(0.2303 \log \frac{y}{x}\)
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
For first order reaction:
\[
k = \frac{1}{t_2 - t_1} \ln \frac{[A]_{t_1}}{[A]_{t_2}} = \frac{1}{10} \ln \frac{x}{y}
\]
Converting to log base 10:
\[
k = \frac{2.303}{10} \log \frac{x}{y} = 0.2303 \log \frac{x}{y}
\]
Was this answer helpful?
0
0
Top Questions on Chemical Kinetics
- For a first-order reaction, the ratio between the time taken to complete $\frac{3}{4}$th of the reaction and time taken to complete half of the reaction is
- AP EAPCET - 2025
- Chemistry
- Chemical Kinetics
- The following equation is obtained for a first order reaction at 300 K.
$\log_{10} \frac{k}{A} = 0.00174$ What is the activation energy (in J mol\(^{-1}\)) of the reaction ?
(R = 8.314 J mol\(^{-1}\) K\(^{-1}\))- AP EAPCET - 2025
- Chemistry
- Chemical Kinetics
- $R \to P$ is a first-order reaction. The concentration of R changed from 0.04 to 0.03 mol L$^{-1}$ in 40 minutes. What is the average velocity of the reaction in mol L$^{-1}$ s$^{-1}$?
- AP EAPCET - 2025
- Chemistry
- Chemical Kinetics
- At T(K), the following equation is obtained for a first order reaction.
\[ \log \frac{k}{A} = -\frac{x}{T} \] The activation energy for this reaction is equal to (R = gas constant)- AP EAPCET - 2025
- Chemistry
- Chemical Kinetics
- A → P is a first order reaction. The reaction was started at 10.00 AM. At 10.10 AM, the concentration of A was x mol L-1. At 10.20 AM, the concentration of A was y mol L-1. The half life (in min) of the reaction is equal to:
- AP EAPCET - 2025
- Chemistry
- Chemical Kinetics
View More Questions
Questions Asked in AP EAPCET exam
- If $x>\sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x$, then the coefficient of $x^{-8}$ is
- AP EAPCET - 2025
- Trigonometric Identities
- If $-\frac{2}{3}<x<\frac{2}{3}$, then the value of the $5^{th}$ term in the expansion of $\frac{1}{\sqrt{2-3x}}$ when $x = \frac{1}{2}$ is
- AP EAPCET - 2025
- Number System
- If the coefficients of $x^{10}$ and $x^{11}$ in the expansion of $(1 + \alpha x + \beta x^2)(1+x)^{11}$ are 396 and 144 respectively, then $\alpha^2 + \beta^2 =$
- AP EAPCET - 2025
- Combinatorics
- If \(1, \omega, \omega^2\) are the cube roots of unity, then \( \left(1\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right) + 2\left(3+\frac{1}{\omega}\right)\left(3+\frac{1}{\omega^2}\right) + 3\left(4+\frac{1}{\omega}\right)\left(4+\frac{1}{\omega^2}\right) + \dots + 10 \text{ terms}\right) = \)
- AP EAPCET - 2025
- Complex numbers
- Evaluate the infinite series: \[ \left|\begin{array}{ccc} 2 & 1 & \frac{1}{3} \\ 3 & 1 & 1 \\ \end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{3} & \frac{1}{2} \\ 3 & 1 & 1 \\ \end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{4} & \frac{1}{9} \\ 3 & 1 & 1 \\ \end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{4} & \frac{1}{27} \\ 3 & 1 & 1 \\ \end{array}\right| + \cdots = ? \]
- AP EAPCET - 2025
- Complex numbers
View More Questions
