For a first order reaction A(g) to 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t and P_(0) and P_(t) respectively, Initially, only A is present with concentration [A]_(0) and t_(1//3) is the required for the partial pressure of A to reach 1//3rd of its initial value. The correct option(s) is are (Assume that all these gases behave as ideal gases)

For a first order reaction A(g) to 2B(g) + C(g) at constant volume and

1.	For a first order reaction A(g) to 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t and P_(0) and P_(t) respectively, Initially, only A is present with concentration [A]_(0) and t_(1//3) is the required for the partial pressure of A to reach 1//3rd of its initial value. The correct option(s) is are (Assume that all these gases behave as ideal gases)
Answer» Solution :(a,d) `A(g) underset("ORDER")OVERSET("First") to2B(g) + C(g)` time =0 `P_(0)` For first order reaction, `t=1/k In P_(0)/(P_(0)-x)` `t=1/k In (P_(0))/(P_(0) - (P_(t)-P_(0))/2)` `=1/k In (2P_(0))/(2P_(0) - P_(t) + P_(0))` `kt = In (2P_(0))/(3P_(0)-P_(t))` `= In 2P_(0) - In(3P_(0)-P_(t))` or `In (3P_(0)-P_(t)) = In 2P_(0)-kt` Graph between in `(3P_(0)-P_(t)) = In 2P_(0)-kt` Graph between in `(3P_(0)-P_(t))` Vs t is a straight line with NEGATIVE slope `therefore` Graph(a) is a correct option. Since rate constant (k) is independent of initial concentration `therefore` graph (d) is also a correct option.