How much time is required for complete decomposition of 4 moles of water using 4 ampere?

Options

A. `3.86xx10^5`sec
B. `1.93xx10^5`sec
C. 96500sec
D. 48250sec

Answer

To find the time required for the complete decomposition of 4 moles of water using a current of 4 amperes, we can follow these steps:

### Step 1: Determine the number of electrons involved
The decomposition of water can be represented by the reaction:
\[ 2H_2O \rightarrow 4H^+ + O_2 + 4e^- \]
From this reaction, we see that 2 moles of water produce 4 moles of electrons (4e⁻). Therefore, for 4 moles of water:
\[ 4 \text{ moles of } H_2O \rightarrow 8H^+ + O_2 + 8e^- \]
This means that 4 moles of water will produce 8 moles of electrons.

### Step 2: Calculate the total charge (Q)
We know that 1 mole of electrons corresponds to 96500 coulombs (Faraday's constant). Thus, for 8 moles of electrons:
\[ Q = 8 \text{ moles} \times 96500 \text{ C/mole} = 772000 \text{ C} \]

### Step 3: Use the formula relating charge, current, and time
The relationship between charge (Q), current (I), and time (t) is given by:
\[ Q = I \times t \]
We can rearrange this to find time:
\[ t = \frac{Q}{I} \]

### Step 4: Substitute the values into the equation
Substituting the values we have:
- \( Q = 772000 \text{ C} \)
- \( I = 4 \text{ A} \)

Thus,
\[ t = \frac{772000 \text{ C}}{4 \text{ A}} = 193000 \text{ seconds} \]

### Step 5: Convert time into a more understandable unit (if necessary)
To convert seconds into hours, we can divide by 3600 (the number of seconds in an hour):
\[ t = \frac{193000 \text{ seconds}}{3600 \text{ seconds/hour}} \approx 53.61 \text{ hours} \]

### Final Answer
The time required for the complete decomposition of 4 moles of water using 4 amperes is approximately **193000 seconds** or **53.61 hours**.

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