Integrated rate equations

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 :

In the reaction,
$$\ce{P + Q -> R + S}$$
the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\dfrac{[t_{1/8}]}{[t_{1/10}]}\times10$? (take $\log_{10}2=0.3$)

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

The concentration of R in the reaction R $\rightarrow$ P was measured as a function of time and the following data is obtained:

[R] (molar)	1.0	0.75	0.40	0.10
t (min.)	0.0	0.05	0.12	0.18

The order of the reaction is