In a zero order reaction , 20% of the reaction complete in 10 s. How much time it will take to complete 50% of the reaction ?

Options

A. 20 s
B. 25 s
C. 30 s
D. 40 s

Answer

To solve the problem step by step, we will use the principles of zero-order kinetics.

### Step 1: Understand the Zero-Order Reaction
In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The rate equation can be expressed as:
\[ [A]_t = [A]_0 - kt \]
where:
- \([A]_t\) = concentration of the reactant at time \(t\)
- \([A]_0\) = initial concentration of the reactant
- \(k\) = rate constant
- \(t\) = time

### Step 2: Determine the Rate Constant \(k\)
Given that 20% of the reaction is complete in 10 seconds, we can express this mathematically. If we assume the initial concentration \([A]_0\) is 100 units, then 20% completion means:
\[ [A]_t = 100 - 20 = 80 \]
Using the formula for zero-order reactions:
\[ 80 = 100 - k(10) \]
Rearranging gives:
\[ k(10) = 100 - 80 \]
\[ k(10) = 20 \]
\[ k = \frac{20}{10} = 2 \, \text{units/time} \]

### Step 3: Calculate Time for 50% Completion
Next, we need to find out how long it will take to complete 50% of the reaction. For 50% completion, the concentration will be:
\[ [A]_t = 100 - 50 = 50 \]
Using the zero-order equation again:
\[ 50 = 100 - kt_{50\%} \]
Rearranging gives:
\[ kt_{50\%} = 100 - 50 \]
\[ kt_{50\%} = 50 \]
Substituting \(k = 2\):
\[ 2t_{50\%} = 50 \]
\[ t_{50\%} = \frac{50}{2} = 25 \, \text{seconds} \]

### Conclusion
The time required to complete 50% of the reaction is **25 seconds**.

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