= Solution
Let $t_i=\mathbb E_i[X]$. Time is accumulated in each transient chamber before the next move; arrival at either terminal state ends the trial immediately. Thus $t_0=t_k=0$. A block of $k$ identical rightward choices forces absorption, so the <geometric tail bound from a uniform escape probability> gives finite <expected value> of $X$.
<First-step analysis> now gives
$$
t_i=1+\tfrac12t_{i-1}+\tfrac12t_{i+1},\qquad1\leq i\leq k-1.
$$
The quadratic $t_i=i(k-i)$ has the correct boundary values and second difference $t_{i+1}-2t_i+t_{i-1}=-2$, so it solves this system. The difference of two solutions would be linear in $i$ with zero values at both endpoints, hence zero. Therefore
$$
\boxed{\mathbb E_i[X]=i(k-i)\text{ minutes}.}
$$
This is the <expected duration of symmetric gambler's ruin>. It includes the first minute in the starting transient chamber and does not add a minute after absorption.
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