A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of CO2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then KP is : (1) 0.18 atm (2) 1.8 atm (3) 0.3 atm (4) 3 atm.

📚 Key Concepts Needed

1. Gas-phase equilibrium constant $K_P$

   For a gaseous reaction
   $aA + bB \rightleftharpoons cC + dD$
   the pressure form of the equilibrium constant is
   $K_P = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$
   where $P_X$ are partial pressures at equilibrium.

2. Solids & liquids are ignored in $K_P$ expressions because their activities are essentially 1.

3. Total pressure in a mixture of gases equals the sum of their partial pressures:
   $P_{total} = \sum P_i$

4. ICE or RICE table method (Initial, Change, Equilibrium) is a systematic way to relate initial conditions, changes caused by the reaction, and equilibrium values.

5. Stoichiometry link: In $C + CO_2 \to 2CO$, every 1 mole (or 1 atm in an ideal gas sense) of $CO_2$ consumed produces 2 moles (atm) of $CO$.

📝 Logical Chain to Solve

1. Write the balanced equation and set up the RICE table for partial pressures.
2. Assign a variable $x$ for the amount of $CO_2$ that reacts.
3. Apply the stoichiometric ratios to find how much $CO$ forms (twice $x$).
4. Use the given total pressure at equilibrium to solve for $x$.
5. Compute the equilibrium partial pressures of $CO$ and $CO_2$.
6. Plug them into the $K_P$ formula (remember graphite is a solid, so it is not included).
7. Select the matching option from the choices.

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Step-by-Step Solution

Balanced reaction:
$C\,(s) + CO_2\,(g) \rightleftharpoons 2\,CO\,(g)$

Total equilibrium pressure:
$P_{total} = (0.5 - x) + 2x = 0.5 + x$

Given: $P_{total} = 0.8\,\text{atm}$

So $0.5 + x = 0.8 \;\Rightarrow\; x = 0.3$

Therefore

$P_{CO_2}^{eq} = 0.5 - 0.3 = 0.2\,\text{atm}$

$P_{CO}^{eq} = 2x = 2(0.3) = 0.6\,\text{atm}$

Now compute $K_P$ (solid $C$ omitted):

$K_P = \frac{(P_{CO})^{2}}{P_{CO_2}}$

Substitute values:

$K_P = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8\,\text{atm}$

Answer: oxed{1.8\,\text{atm}} (Option 2).