The initial concentration of cane sugar in presence of an acid was reduced from 0.20 to 0.10 M in 5 hours and to 0.05 M in 10 hours, value of K ? (in `hr^(-1)`)

Options

A. 0.693
B. 1.386
C. 0.1386
D. 3.465

Answer

To solve the problem, we need to determine the rate constant (K) for the reaction based on the given concentration changes of cane sugar over time. The steps are as follows:

### Step 1: Understand the Reaction
The concentration of cane sugar decreases over time in the presence of an acid. We are given the following data:
- Initial concentration (C₀) = 0.20 M
- Concentration after 5 hours (C₁) = 0.10 M
- Concentration after 10 hours (C₂) = 0.05 M

### Step 2: Determine the Half-Life
The half-life (t₁/₂) is the time taken for the concentration of a reactant to decrease to half of its initial value.

From 0.20 M to 0.10 M (5 hours), we see that the concentration is halved. Thus, the half-life for this reaction is:
- t₁/₂ = 5 hours

### Step 3: Use the First-Order Reaction Formula
For a first-order reaction, the relationship between half-life (t₁/₂) and the rate constant (K) is given by the formula:
\[ t_{1/2} = \frac{0.693}{K} \]

### Step 4: Rearrange the Formula to Solve for K
We can rearrange the formula to find K:
\[ K = \frac{0.693}{t_{1/2}} \]

### Step 5: Substitute the Half-Life Value
Now, substitute the half-life value into the equation:
\[ K = \frac{0.693}{5 \text{ hours}} \]

### Step 6: Calculate K
Now, perform the calculation:
\[ K = \frac{0.693}{5} = 0.1386 \text{ hr}^{-1} \]

### Conclusion
The value of the rate constant (K) is 0.1386 hr⁻¹.

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