For a zero order reaction,
`K=1xx10^(-3)"mol L"^(-1)s^(-1)`
If initial concentration of the reactant is `1.0"mol L"^(-1)` , the concentration after 10 minutes would be

Options

A. `1xx10^(-2)"mol L"^(-1)`
B. `0.6 "mol L"^(-1)`
C. `0.4 "mol L"^(-1)`
D. `1.0 "mol L"^(-1)`

Answer

To solve the problem, we will use the formula for a zero-order reaction. The concentration of the reactant at any time \( t \) can be calculated using the following equation:

\[ [A]_t = [A]_0 - kt \]

Where:
- \([A]_t\) is the concentration of the reactant at time \( t \).
- \([A]_0\) is the initial concentration of the reactant.
- \( k \) is the rate constant.
- \( t \) is the time.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Rate constant, \( k = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \)
- Initial concentration, \( [A]_0 = 1.0 \, \text{mol L}^{-1} \)
- Time, \( t = 10 \, \text{minutes} = 10 \times 60 = 600 \, \text{seconds} \)

2. **Substitute the Values into the Zero-Order Reaction Formula:**
\[ [A]_t = [A]_0 - kt \]
Substituting the known values:
\[ [A]_t = 1.0 \, \text{mol L}^{-1} - (1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1}) \times (600 \, \text{s}) \]

3. **Calculate \( kt \):**
\[ kt = (1 \times 10^{-3}) \times 600 = 0.6 \, \text{mol L}^{-1} \]

4. **Calculate \([A]_t\):**
\[ [A]_t = 1.0 - 0.6 = 0.4 \, \text{mol L}^{-1} \]

5. **Final Result:**
The concentration of the reactant after 10 minutes is:
\[ [A]_t = 0.4 \, \text{mol L}^{-1} \]

### Conclusion:
The concentration after 10 minutes would be \( 0.4 \, \text{mol L}^{-1} \).

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