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What is the bond length between nitrogen and oxygen atoms in NO molecule? Given the bond energy of NO is 602 kJ/mol and the reduced mass of the NO molecule is 14.8 amu.

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To calculate the bond length between nitrogen and oxygen atoms in the NO molecule, we can use the Morse potential equation, which relates bond energy, reduced mass, and bond length. The Morse potential equation is:V r  = D 1 - e^-a r - r0   ^2where V r  is the potential energy, D is the bond energy, a is a constant, r is the bond length, and r0 is the equilibrium bond length.We can also use the vibrational frequency equation: =  1/ 2   * sqrt k/ where  is the vibrational frequency, k is the force constant, and  is the reduced mass.First, we need to find the force constant  k  using the bond energy  D  and reduced mass   . We can do this by rearranging the vibrational frequency equation:k =  2 ^2 * We are given the bond energy  D  as 602 kJ/mol, which we need to convert to J/mol:602 kJ/mol *  1000 J/1 kJ  = 602000 J/molNow, we need to find the vibrational frequency   . We can estimate this using the bond energy and the reduced mass:  sqrt D/ We are given the reduced mass    as 14.8 amu, which we need to convert to kg:14.8 amu *  1.66054 x 10^-27 kg/1 amu  = 2.4576 x 10^-26 kgNow, we can find the vibrational frequency   :  sqrt  602000 J/mol  /  2.4576 x 10^-26 kg    1.564 x 10^13 HzNow, we can find the force constant  k :k =  2 1.564 x 10^13 Hz  ^2 *  2.4576 x 10^-26 kg k  9.614 x 10^2 N/mNow, we can use the Morse potential equation to find the equilibrium bond length  r0 . We need to find the value of the constant  a  first. We can estimate this using the force constant  k  and the bond energy  D :a  sqrt k /  2D  a  sqrt  9.614 x 10^2 N/m  /  2 * 602000 J/mol  a  1.131 x 10^-10 m^-1Now, we can find the equilibrium bond length  r0  using the Morse potential equation. We can rearrange the equation to solve for r0:r0 = r -  1/a  * ln 1 - sqrt V r  / D  Since we want to find the bond length at the minimum potential energy, we can set V r  to 0:r0 = r -  1/a  * ln 1 - sqrt 0 / D  r0 = r -  1/ 1.131 x 10^-10 m^-1   * ln 1 r0 = rThus, the bond length between nitrogen and oxygen atoms in the NO molecule is approximately equal to the equilibrium bond length  r0 , which is approximately 1.131 x 10^-10 m or 1.131 .
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