 Solution

Use the same standard pressure and temperature, the same bond length, and the same electronic and vibrational-ground-state approximations. The translational partition function scales as mass to the power $3/2$, giving the factor $(47/44{)}^{3/2}$. For the original isotopologue, $I=32m_u{R}^2$ and the rotational symmetry number is two. For the isotope-substituted molecule, the two oxygen ends are distinguishable, so the symmetry number is one. HenceThe change in standard molar entropy issoThis uses a convention omitting a separate nuclear-spin entropy. If nuclear-spin degeneracies are instead included explicitly, the spin-$1/2$ carbon-13 nucleus adds a further $R\log 2$ relative to the spin-zero original isotopologue; that convention must be stated rather than silently mixed with the quoted partition functions.