Question:medium

For a first order reaction ($R \rightarrow P$), a plot of $\ln \frac{[R]_0}{[R]}$ on y-axis and time on x-axis gave a straight line passing through the origin. The slope of straight line is 'x'. The rate constant of this reaction is

Show Hint

Be careful with log bases. - Plot of $\ln(R_0/R)$ vs $t$: Slope = $k$. - Plot of $\log_{10}(R_0/R)$ vs $t$: Slope = $k/2.303$.

Updated On: Mar 31, 2026

$2.303 x$
$x^{-1}$
$x$
$2.303$

Show Solution

The Correct Option is C

Solution and Explanation

Was this answer helpful?

0

Top Questions on Chemical Kinetics

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra