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 integrated rate law for a zero-order reaction. The steps are as follows:

### Step 1: Understand the Zero-Order Reaction
For a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactant. The integrated rate law for a zero-order reaction is given by:

\[ [A] = [A_0] - kt \]

where:
- \([A]\) = concentration of the reactant at time \(t\)
- \([A_0]\) = initial concentration of the reactant
- \(k\) = rate constant
- \(t\) = time

### Step 2: Identify Given Values
From the problem, we have:
- \(k = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1}\)
- \([A_0] = 1.0 \, \text{mol L}^{-1}\)
- \(t = 10 \, \text{minutes}\)

### Step 3: Convert Time to Seconds
Since the rate constant \(k\) is given in terms of seconds, we need to convert the time from minutes to seconds:

\[ t = 10 \, \text{minutes} \times 60 \, \text{seconds/minute} = 600 \, \text{seconds} \]

### Step 4: Substitute Values into the Integrated Rate Law
Now we can substitute the values into the integrated rate law equation:

\[ [A] = [A_0] - kt \]

Substituting the known values:

\[ [A] = 1.0 \, \text{mol L}^{-1} - (1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1})(600 \, \text{s}) \]

### Step 5: Calculate the Change in Concentration
Now calculate \(kt\):

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

### Step 6: Calculate the Final Concentration
Now substitute back to find \([A]\):

\[ [A] = 1.0 \, \text{mol L}^{-1} - 0.6 \, \text{mol L}^{-1} = 0.4 \, \text{mol L}^{-1} \]

### Conclusion
The concentration of the reactant after 10 minutes is:

\[ \boxed{0.4 \, \text{mol L}^{-1}} \]

---