For the formation of `NH_4HS`
`NH_3(g) +H_2S(g) hArr NH_4HS(s)`
The energy for formation is `DeltaH=-85 kcal`. Find the maximum rate constant of backward reaction at `27^@C` . The energy difference between reactant and threshold energy is `+105 kJ(K_b=5.4 xx 10^(-3) "mol litre"^(-1) sec ^(-1)`)

Explanation

To solve the problem, we need to find the maximum rate constant of the backward reaction for the formation of `NH4HS` from `NH3` and `H2S`. The reaction is given as:

\[ \text{NH}_3(g) + \text{H}_2\text{S}(g) \rightleftharpoons \text{NH}_4\text{HS}(s) \]

### Step 1: Identify the given data
- The enthalpy change for the formation of `NH4HS` is given as:
\[ \Delta H = -85 \text{ kcal} \]
- The activation energy for the forward reaction (from reactants to products) is given as:
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