In an elementary reaction `A(g)+2B(g) to C(g)` the initial pressure of A and B are `P_(A)=0.40` atm and `P_(B)`=0.60 atm respectively. After time T, if pressure of C is observed 0.1 atm, then find the value of `(r_(i) (" initial rate of reaction"))/(r_(t)(" rate of reaction after time t"))`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ A(g) + 2B(g) \rightarrow C(g) \]
The initial pressures are:
- \( P_A = 0.40 \, \text{atm} \)
- \( P_B = 0.60 \, \text{atm} \)

### Step 2: Determine Changes in Pressure
After time \( T \), the pressure of \( C \) is observed to be \( 0.1 \, \text{atm} \). From the stoichiometry of the reaction, for every 1 mole of \( A \) consumed, 2 moles of \( B \) are consumed, and 1 mole of \( C \) is produced.

Let \( x \) be the change in pressure of \( A \) and \( C \):
- The pressure of \( C \) formed is \( 0.1 \, \text{atm} \), so \( x = 0.1 \, \text{atm} \).
- The pressure of \( A \) after time \( T \):
\[ P_A = 0.40 - 0.1 = 0.30 \, \text{atm} \]
- The pressure of \( B \) after time \( T \):
\[ P_B = 0.60 - 2 \times 0.1 = 0.60 - 0.2 = 0.40 \, \text{atm} \]

### Step 3: Calculate Initial Rate of Reaction
The initial rate of reaction \( r_i \) can be expressed as:
\[ r_i = k P_{A0} P_{B0}^2 \]
Substituting the initial pressures:
\[ r_i = k (0.40) (0.60)^2 \]
Calculating this:
\[ r_i = k (0.40) (0.36) = k (0.144) \]

### Step 4: Calculate Rate of Reaction After Time \( T \)
The rate of reaction after time \( T \) \( r_t \) is given by:
\[ r_t = k P_A P_B^2 \]
Substituting the pressures after time \( T \):
\[ r_t = k (0.30) (0.40)^2 \]
Calculating this:
\[ r_t = k (0.30) (0.16) = k (0.048) \]

### Step 5: Find the Ratio of Initial Rate to Rate After Time \( T \)
Now, we need to find the ratio \( \frac{r_i}{r_t} \):
\[ \frac{r_i}{r_t} = \frac{k (0.144)}{k (0.048)} = \frac{0.144}{0.048} = 3 \]

### Final Answer
Thus, the value of \( \frac{r_i}{r_t} \) is \( 3 \).

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