Determine which of the following reactions at constant pressure represents surroundings that do work on the system:
I. `4NH_(3)(g)+7O_(2)(g)rarr4NO_(2)(g)+6H_(2)O(g)`
II. `CO(g)+2H_(2)(g)rarrCH_(3)OH(l)`
III. `C(s,"graphite")+H_(2)O(g)rarrCO(g)+H_(2)(g)`
IV. `H_(2)OrarrH_(2)O(l)`

Options

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

Answer

To determine which of the given reactions at constant pressure represents surroundings that do work on the system, we need to analyze the change in the number of moles of gaseous reactants and products. The work done on the system can be evaluated using the formula:

\[ W = -P \Delta V \]

Where \( \Delta V \) is the change in volume, which can be related to the change in the number of moles of gas (\( \Delta n_G \)):

\[ \Delta n_G = n_{products} - n_{reactants} \]

If \( \Delta n_G < 0 \) (i.e., the number of moles of gas decreases), then the work done is positive, indicating that work is done on the system.

Let's analyze each reaction step by step:

### Reaction I:
\[ 4NH_3(g) + 7O_2(g) \rightarrow 4NO_2(g) + 6H_2O(g) \]

- **Reactants:** 4 moles of \( NH_3 \) + 7 moles of \( O_2 \) = 11 moles of gas
- **Products:** 4 moles of \( NO_2 \) + 6 moles of \( H_2O \) = 10 moles of gas
- **Change in moles:** \( \Delta n_G = 10 - 11 = -1 \)

Since \( \Delta n_G < 0 \), work is done on the system.

### Reaction II:
\[ CO(g) + 2H_2(g) \rightarrow CH_3OH(l) \]

- **Reactants:** 1 mole of \( CO \) + 2 moles of \( H_2 \) = 3 moles of gas
- **Products:** 0 moles of gas (since \( CH_3OH \) is a liquid)
- **Change in moles:** \( \Delta n_G = 0 - 3 = -3 \)

Since \( \Delta n_G < 0 \), work is done on the system.

### Reaction III:
\[ C(s, graphite) + H_2O(g) \rightarrow CO(g) + H_2(g) \]

- **Reactants:** 1 mole of \( H_2O \) = 1 mole of gas
- **Products:** 1 mole of \( CO \) + 1 mole of \( H_2 \) = 2 moles of gas
- **Change in moles:** \( \Delta n_G = 2 - 1 = +1 \)

Since \( \Delta n_G > 0 \), work is done by the system.

### Reaction IV:
\[ H_2O(g) \rightarrow H_2O(l) \]

- **Reactants:** 1 mole of \( H_2O \) (gas)
- **Products:** 0 moles of gas (since \( H_2O \) is a liquid)
- **Change in moles:** \( \Delta n_G = 0 - 1 = -1 \)

Since \( \Delta n_G < 0 \), work is done on the system.

### Summary of Results:
- **Reactions that do work on the system:** I, II, IV
- **Reaction that does not do work on the system:** III

### Final Answer:
The reactions that represent surroundings that do work on the system are I, II, and IV.

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