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0.25 mol of CO taken in a 1.5 L flask is...

`0.25` mol of `CO` taken in a `1.5 L` flask is maintained at `500 K` along with a catalyst so that the following reaction can take place:
`CO(g)+H_(2)(g)hArrCH_(3)OH(g)` .
Hydrogen is introduced until the total pressure of system is `8.2 atm`, at equilibrium, and `0.1 mol` of methanol is formed. Calculate
a. `K_(p)` and `K_(c )`
b. The final pressure if the same amount of `CO` and `H_(2)` as before are used but no catalyst so that the reaction does take place.

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To solve the problem step by step, we will break it down into two parts: calculating \( K_p \) and \( K_c \), and then finding the final pressure without a catalyst. ### Part A: Calculation of \( K_p \) and \( K_c \) 1. **Write the balanced chemical equation:** \[ \text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \] ...
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