At room temperaturee, for the reaction
`NH_(4)SH_((s))hArrNH_(3(s))+H_(2)S_((g))`

Options

A. `K_(p)=K_(c)`
B. `K_(p)gtK_(c)`
C. `K_(p)ltK_(c)`
D. `K_(p)and K_(c)` do not relate

Answer

To solve the problem regarding the relationship between \( K_p \) and \( K_c \) for the reaction:

\[ NH_4SH_{(s)} \rightleftharpoons NH_3_{(s)} + H_2S_{(g)} \]

we need to follow these steps:

### Step 1: Identify the Reaction Components
In the given reaction, we have:
- Reactants: \( NH_4SH \) (solid)
- Products: \( NH_3 \) (solid) and \( H_2S \) (gas)

### Step 2: Determine the Change in Moles of Gas (\( \Delta N_g \))
The change in moles of gas (\( \Delta N_g \)) is calculated as follows:
- Moles of gaseous products: 1 (from \( H_2S \))
- Moles of gaseous reactants: 0 (since \( NH_4SH \) is solid and does not contribute to the gas phase)

Thus,

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 0 = 1 \]

### Step 3: Relate \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:

\[ K_p = K_c (RT)^{\Delta N_g} \]

Where:
- \( R \) is the universal gas constant
- \( T \) is the temperature in Kelvin
- \( \Delta N_g \) is the change in moles of gas

### Step 4: Analyze the Sign of \( \Delta N_g \)
Since we found that \( \Delta N_g = 1 \), which is greater than zero, we can conclude:

- When \( \Delta N_g > 0 \), it implies that \( K_p > K_c \).

### Conclusion
For the reaction \( NH_4SH_{(s)} \rightleftharpoons NH_3_{(s)} + H_2S_{(g)} \), since \( \Delta N_g > 0 \), we conclude that:

\[ K_p > K_c \]

Thus, the answer is that \( K_p \) is greater than \( K_c \) for this reaction.

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