Question

In the reaction $P_4S_3 \rightarrow P_2O_5 + SO_2$ Equivalent weight of $P_4S_3$ (M = molar mass of $P_4S_3$) is

• A. M/14
• B. M/38
• C. M/18
• D. M/32

Answer 1

Answer: (D)

M/32

Answer 2

The balanced chemical equation for the reaction is: $P_4S_3 + 8 O_2 \rightarrow 2 P_2O_5 + 3 SO_2$

To find the equivalent weight of $P_4S_3$, we need to determine its n-factor.

1. Oxidation states in $P_4S_3$: Assuming the oxidation state of S is -2: $4 \times OS(P) + 3 \times (-2) = 0$ $4 \times OS(P) = +6$ $OS(P) = +1.5$ (average)

2. Oxidation states in products: In $P_2O_5$: $2 \times OS(P) + 5 \times (-2) = 0 \implies OS(P) = +5$. In $SO_2$: $OS(S) + 2 \times (-2) = 0 \implies OS(S) = +4$.

3. Change in oxidation states:

   • Phosphorus (P): Total change for 4 P atoms = $4 \times (+5) - 4 \times (+1.5) = 20 - 6 = +14$.
   • Sulfur (S): Total change for 3 S atoms = $3 \times (+4) - 3 \times (-2) = 12 - (-6) = +18$.

4. n-factor: The total change in oxidation states for $P_4S_3$ is the sum of the changes for P and S. n-factor = $|+14| + |+18| = 14 + 18 = 32$.

5. Equivalent Weight: Equivalent Weight = $\frac{\text{Molar Mass}}{\text{n-factor}} = \frac{M}{32}$.