Answer
$t = 150\text{ s}$
Work Step by Step
Take note that the integrated rate law for first-order reactions is,
$ln \bigg ( \dfrac{[A]}{[A]_0} \bigg ) = -kt$
Thus to solve for the constant k,
$ln \bigg ( \dfrac{0.250 \ mol/L}{1.00 \ mol/L} \bigg ) = -k (120 \ s), \ k = 0.0116 s^{-1}$
Then, to solve for the time it takes for 2.00 mol/L of PH$_3$ to decrease to 0.350 mol/L is,
$ln \bigg ( \dfrac{0.350 \ mol/L}{2.00 \ mol/L} \bigg ) = -0.0116 s^{-1} \times t, \ t = 150 \ s$
