Decomposition of phosphine `(PH_(3))` at `120^(@)` C proceeds according to the equation
`4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)`
It was found that the reaction follows the rate equation, Rate = `k[PH_(3)]`
The half life period of `PH_(3)` is 37.9 s at `120^(@)`C.
i) How much time will be required for `3//4` of `PH_(3)` to decompose?
ii) What function of the original amount of `PH_(3)` will remain undecomposed after 1 minute?

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part i: Time required for \( \frac{3}{4} \) of \( PH_3 \) to decompose

1. **Identify the reaction order and half-life formula**:
The decomposition of phosphine follows a first-order reaction, as indicated by the rate equation \( \text{Rate} = k[PH_3] \). The half-life \( t_{1/2} \) for a first-order reaction is given by the formula:
\[
t_{1/2} = \frac{0.693}{k}
...