Half-life

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

(Use: $\ln 2 = 0.693$)

Which of the following plots is(are) correct for the given reaction?

($[\mathbf P]_0$ is the initial concentration of P)

For a first order reaction $\mathrm{A(g)} \to \mathrm{2B(g)} + \mathrm{C(g)}$ at constant volume and 300 K, the total pressure at the beginning ($t = 0$) and at time $t$ are $P_0$ and $P_t$, respectively. Initially, only A is present with concentration $[\mathrm{A}]_0$, and $t_{1/3}$ is the time required for the partial pressure of A to reach 1/3$^{\text{rd}}$ of its initial value. The correct option(s) is (are)

(Assume that all these gases behave as ideal gases)

Decomposition of $\ce{H2O2}$ follows a first order reaction. In fifty minutes the concentration of $\ce{H2O2}$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $\ce{H2O2}$ reaches 0.05 M, the rate of formation of $\ce{O2}$ will be :

The reaction
$$\ce{2N2O5(g) -> 4NO2(g) + O2(g)}$$
follows first order kinetics. The pressure of a vessel containing only $\ce{N2O5}$ was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant) :

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be :

For the first order reaction
$$\ce{2N2O5(g) -> 4NO2(g) + O2(g)}$$

Under the same reaction conditions, initial concentration of 1.386 mol dm⁻³ of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio $\left(\dfrac{k_1}{k_0}\right)$ of the rate constants for first order $(k_1)$ and zero order $(k_0)$ of the reactions is