JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)

Solid carbon, CaO and CaCO\(_3\) are mixed and allowed to attain equilibrium at T K.
\(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{p_1} = 0.08\) atm
\(\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \quad K_{p_2} = 2\) atm
The partial pressure of CO is ________ \(\times 10^{-1}\) atm

Verified Solution

Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

For the first equilibrium: \(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\) \(K_{p_1} = P_{\text{CO}_2} = 0.08 \text{ atm}\) For the second equilibrium: \(\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

The reaction between \(X\) and \(Y\) is first order with respect to \(X\) and zero order with respect to \(Y\).

Experiment	\(\frac{[ X ]}{ mol \;L ^{-1}}\)	\(\frac{[ Y ]}{ mol\; L ^{-1}}\)	\(\frac{\text { Initial rate }}{ mol\; L ^{-1}\; min ^{-1}}\)
\(I\)	\(0.1\)	\(0.1\)	\(2 \times 10^{-3}\)
\(II\)	\(0.2\)	\(0.2\)	\(4 \times 10^{-3}\)
\(III\)	\(0.4\)	\(0.4\)	\(M \times 10^{-3}\)
\(IV\)	\(0.1\)	\(0.2\)	\(2 \times 10^{-3}\)

Examine the data of table and calculate ratio of numerical value of \(M\)

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