For the following reaction

the rate of reaction is $$\frac{d[P]}{dt} = k[X]$$. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are)
(Use: ln 2 = 0.693)

The rate law is given as $$\frac{d[P]}{dt}=k[X]$$. This is a first-order rate law with respect to $$X$$ and it is written for the formation of one mole of $$P$$.

Step 1 : Deduce the stoichiometric coefficients.
Initially: $$[X]_0 = 2\;{\rm M},\;[Y]_0 = 1\;{\rm M}$$ in a 1 L solution.
After 50 s: $$[Y] = 0.5\;{\rm M}$$, so $$0.5\;{\rm mol\;L^{-1}}$$ of $$Y$$ has reacted.

Let the balanced equation be $$\nu X + Y \rightarrow P$$. For every mole of $$Y$$ consumed, $$\nu$$ moles of $$X$$ are consumed and one mole of $$P$$ is formed.

Since 0.5 M of $$Y$$ is consumed in 50 s, consumption of $$X$$ is $$\nu \times 0.5 \;{\rm M}$$.
Thus $$[X]_{50\,{\rm s}} = 2 - 0.5\nu \;{\rm M}$$ $$-(1)$$

Step 2 : Use the integrated first-order equation.
For a first-order process, $$\ln\!\left(\frac{[X]_0}{[X]}\right)=kt$$ $$-(2)$$

Substituting $$[X]_0 = 2\;{\rm M},\;[X]=2-0.5\nu \;{\rm M},\;t = 50\;{\rm s}$$ in $$(2)$$ gives $$\ln\!\left(\frac{2}{\,2-0.5\nu\,}\right)=k(50)$$ $$-(3)$$

The problem data are such that $$\ln 2 = 0.693$$ appears useful. If $$[X]_{50\,{\rm s}} = 1\;{\rm M}$$ (i.e. the concentration of $$X$$ exactly halves), then $$2-0.5\nu = 1 \;\Longrightarrow\; \nu = 2$$. Hence the balanced reaction is

$$2X + Y \rightarrow P$$

Step 3 : Evaluate the rate constant, $$k$$.
Putting $$[X]=1\;{\rm M}$$ in $$(2)$$, $$\ln\!\left(\frac{2}{1}\right)=k(50) \;\Longrightarrow\; 0.693 = 50k \;\Longrightarrow\; k = \frac{0.693}{50} = 0.01386\;{\rm s^{-1}}$$ $$-(4)$$

Step 4 : Half-life of $$X$$.
For a first-order reaction, $$t_{1/2}=\frac{\ln 2}{k}$$.
Using $$(4)$$, $$t_{1/2} = \frac{0.693}{0.01386} = 50\;{\rm s}$$ $$-(5)$$

Step 5 : Check each statement.

Option A: proposes $$k = 13.86 \times 10^{-4}\;{\rm s^{-1}} = 1.386 \times 10^{-3}\;{\rm s^{-1}}$$. Actual $$k$$ from $$(4)$$ is $$1.386 \times 10^{-2}\;{\rm s^{-1}}$$, an order of magnitude larger. Option A is incorrect.

Option B: states the half-life of $$X$$ is 50 s. Equation $$(5)$$ confirms this. Option B is correct.

Option C: asks for $$-\frac{d[X]}{dt}$$ at 50 s.
Because the stoichiometric coefficient of $$X$$ is 2, $$-\frac{d[X]}{dt}=2\frac{d[P]}{dt}=2k[X]$$ $$-(6)$$ At 50 s, $$[X]=1\;{\rm M}$$ and $$k=0.01386\;{\rm s^{-1}}$$, so $$-\frac{d[X]}{dt}=2(0.01386)(1)=0.02772\;{\rm mol\,L^{-1}\,s^{-1}} = 27.72 \times 10^{-3}\;{\rm mol\,L^{-1}\,s^{-1}}$$, not $$13.86 \times 10^{-3}$$ as stated. Option C is incorrect.

Option D: To find $$-\frac{d[Y]}{dt}$$ at 100 s, first get $$[X]_{100}$$. From $$(2)$$, $$[X]_{100}=2e^{-k(100)} = 2e^{-1.386}=0.5\;{\rm M}$$. Rate of formation of $$P$$ is $$k[X]=0.01386(0.5)=0.00693\;{\rm mol\,L^{-1}\,s^{-1}}$$. Since one mole of $$Y$$ is consumed per mole of $$P$$ formed, $$-\frac{d[Y]}{dt}=0.00693\;{\rm mol\,L^{-1}\,s^{-1}} = 6.93 \times 10^{-3}\;{\rm mol\,L^{-1}\,s^{-1}}$$, not $$3.46 \times 10^{-3}$$. Option D is incorrect.

Hence, the only correct choice is:
Option B which is: Half-life of X is 50 s.