If the initial concentration is reduced to its $1/4^{th}$ in a zero order reaction,the time taken for half of the reaction to complete

  • A
    Remains same
  • B
    Becomes $4$ times
  • C
    Becomes one-fourth
  • D
    Doubles

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Similar Questions

The rate of a certain reaction depends on concentration according to the equation $-dc/dt = (K_1 C) / (1 + K_2 C)$. What is the order when concentration $(C)$ is very high?

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What is the concentration (in $mol \ L^{-1}$) of the product $B$ after $20 \ s$ in the following reaction? Given that $A \longrightarrow 3B$,rate $= k[A]^0$. The data is provided in the table below:
| Time $(s)$ | Concentration of reactant $A$ $(mol \ L^{-1})$ |
| :--- | :--- |
| $0$ | $0.1$ |
| $15$ | $0.05$ |
| $20$ | $0.1 - x$ |

What is the formula to find the value of $t_{1/2}$ for a Zero Order Reaction?

Mention True $(T)$ and False $(F)$ statements for the following expressions related to a zero-order reaction $R \to P$:
$(i) \ k = \frac{[R]_0 - [R]}{t}$
$(ii) \ k = \frac{[R] - [R]_0}{t}$

In the following reaction $A \to B + C$,the rate constant is $0.001 \, M/sec$. If we start with $1 \, M$ of $A$,the concentrations of $A$ and $B$ after $10$ minutes are respectively:

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