1. Set up the reaction and pressure relation

The reaction is

$$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$$

Initially, only $A$ is present. Let the initial pressure of $A$ be $P_0$.

If fraction decomposed at time $t$ is $\alpha$, then:

• pressure of $A$ left $= P_0(1-\alpha)$
• pressure of $B$ formed $= 2\alpha P_0$
• pressure of $C$ formed $= \alpha P_0$

So total pressure at time $t$ is

$$P_t = P_0(1-\alpha) + 2\alpha P_0 + \alpha P_0 = P_0(1+2\alpha)$$

At completion, $\alpha = 1$, hence

$$P_\infty = P_0(1+2)=3P_0$$

Given:

$$P_\infty = 240\ \text{mm Hg}$$

Therefore,

$$3P_0 = 240 \implies P_0 = 80\ \text{mm Hg}$$

So Option A is correct.

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2. Find extent of reaction at $t=10$ min

Given total pressure at $t=10$ min:

$$P_{10} = 160\ \text{mm Hg}$$

Using

$$P_t = P_0(1+2\alpha)$$

we get

$$160 = 80(1+2\alpha)$$ $$2 = 1+2\alpha$$ $$2\alpha = 1 \implies \alpha = 0.5$$

Thus half of $A$ has decomposed in 10 min.

So partial pressure of $A$ after 10 min is

$$P_A = P_0(1-\alpha)=80(1-0.5)=40\ \text{mm Hg}$$

So Option C is correct.

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3. Find the rate constant

For a first-order reaction,

$$P_A = P_{A0} e^{-kt}$$

At $t=10$ min,

$$40 = 80 e^{-10k}$$ $$e^{-10k} = \frac{1}{2}$$

Taking logarithm,

$$10k = \ln 2$$ $$k = \frac{0.693}{10} = 0.0693\ \text{min}^{-1}$$

So the rate constant is

$$\boxed{0.0693\ \text{min}^{-1}}$$

not $1.693\ \text{min}^{-1}$.

Hence Option D is incorrect.

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4. Check Option B

A first-order reaction approaches completion asymptotically, i.e. theoretically it takes infinite time for complete disappearance of reactant.

So the statement "The reaction never goes to completion" is taken as correct in kinetics language.

Thus Option B is correct.

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5. Final conclusion

The incorrect option is

$$\boxed{\text{D}}$$