Determine which of the following reactions at constant pressure represent surrounding that do work on the system environment :
(P) `4NH_(3)(g) + 7O_(2)(g) to 4 NO_(2) (g) + 6H_(2)O(g)`
(Q) `CO(g)+ 2H_(2)(g) to CH_(3)OH(l)`
(R) `C("s,graphite")+H_(2)O(g) to CO(g)+H_(2)(g)`
(S) `H_(2)O(s) to H_(2)O(l)`

Options

A. III, IV
B. II and III
C. II, IV
D. I and II, IV

Explanation

To determine which reactions represent the surroundings doing work on the system, we need to analyze the change in the number of gaseous moles (ΔNG) for each reaction. The work done on the system by the surroundings is considered positive, and it can be calculated using the formula:

\[ \text{Work} = -P \Delta V = -\Delta N_G RT \]

where:
- \( P \) is the pressure,
- \( \Delta V \) is the change in volume,
- \( \Delta N_G \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

### Step-by-Step Solution:

1. **Reaction (P)**:
\[ 4NH_3(g) + 7O_2(g) \rightarrow 4NO_2(g) + 6H_2O(g) \]
- **Calculate ΔNG**:
- Moles of gaseous products = 4 (NO2) + 6 (H2O) = 10
- Moles of gaseous reactants = 4 (NH3) + 7 (O2) = 11
- ΔNG = 10 - 11 = -1
- **Work Calculation**:
\[ \text{Work} = -(-1)RT = +RT \quad (\text{positive}) \]
- **Conclusion**: Work is done on the system.

2. **Reaction (Q)**:
\[ CO(g) + 2H_2(g) \rightarrow CH_3OH(l) \]
- **Calculate ΔNG**:
- Moles of gaseous products = 0 (CH3OH is liquid)
- Moles of gaseous reactants = 1 (CO) + 2 (H2) = 3
- ΔNG = 0 - 3 = -3
- **Work Calculation**:
\[ \text{Work} = -(-3)RT = +3RT \quad (\text{positive}) \]
- **Conclusion**: Work is done on the system.

3. **Reaction (R)**:
\[ C(s, graphite) + H_2O(g) \rightarrow CO(g) + H_2(g) \]
- **Calculate ΔNG**:
- Moles of gaseous products = 1 (CO) + 1 (H2) = 2
- Moles of gaseous reactants = 1 (H2O) = 1
- ΔNG = 2 - 1 = +1
- **Work Calculation**:
\[ \text{Work} = -(+1)RT = -RT \quad (\text{negative}) \]
- **Conclusion**: Work is done by the system.

4. **Reaction (S)**:
\[ H_2O(s) \rightarrow H_2O(l) \]
- **Calculate ΔNG**:
- Moles of gaseous products = 0 (no gases)
- Moles of gaseous reactants = 0 (no gases)
- ΔNG = 0 - 0 = 0
- **Work Calculation**:
\[ \text{Work} = 0 \quad (\text{no work done}) \]
- **Conclusion**: No work is done.

### Final Answer:
The reactions that represent the surroundings doing work on the system are (P), (Q), and (S). Therefore, the correct options are:

**Correct Options**: P, Q, and S.