V-SMOW is used as a standard for delta2H and delta18O in natural waters wheras V-PDB is rather used as a standard for delta13C contents measurements for dissolved or solid carbonates, sorry the issue is unclear for me...

I am working with 18O KIE (Kinetic isotope effect) for water oxidation. When we measure in mili-Q water,  after collection of O2 and send for analysis. They give the 18O VPDB from where  we can calculate the 18O VPDB to VSMOW. The same experiment I have done in D2O..and I also get 18O VPDB value. Then how I will convert to VSMOW. 

The conversion equation 18O of the values of V-PDB to V-SMOW is:

d18Osmow = d 18Opdb * 1.03086 + 30.86

Thank you Moussa....This equation is valid for H2O. Is this equation is valid for D2O also...???

I think so, there is no reason, it's a mathematical equation

to express the data.

The equation Issam Moussa gave you converts d18O values measured and scale normalized on the VPDB scale to d18O values on the VSMOW scale. For the purpose of this conversion it is immaterial if the measured oxygen originated from water or 2H labelled water (by the way, IUPAC 2005 guidelines on nomenclature have "banned" the us of D - and T - in favour of 2H - and 3H -).  Anyway, the need for the conversion of d18O v. VPDB values to d18O v. VSMOW values arises from the fact that d18O values of carbonate minerals were traditionally linked to the VPDB scale since the scale anchors of the VPDB scale are all carbonate minerals. Abundance values of 18O in all other materials (e.g. phosphates; organic compounds; water) are normalized on the VSMOW scale.

The lab you are sending your water sample for analysis to are using the equilibration method to determine their 18O composition.  This method equilibrates water samples with CO2 gas of a known 18O composition and the measures the change in d18O values after typically 24 hours of equilibration at a constant temperature.  Reference materials used for scale normalization of CO2 stable isotope analysis are NBS 19 and LSVEC (both are carbonates).

 Thank you Prof Wolfram Meier-Augenstein. I measure free 18O VBDP from the D2O bottle which is used for water oxidation. And the value is 602.4. Now I use this D2O solvent for water oxidation to produce oxygen. From the oxygen sample(IRMs study) we got the 18O VPDB value is around 450.6. Now I want to convert both (free D2O and after reaction with D2O) to VSMOW. So what shold I do?

Apply the equation as provided by Issam Moussa to your d18Ovpdb values.

(1.03086 x d18Ovpdb) + 30.86 =  d18Ovsmow

Thank you for response. I apply this equation and measure 18O kinetic isotope effect which gives me very different result than what I got from only H2O. I am affraid about wheter mass and abundance has any effect  on that result. So I am not able to explain what I got from D2O. I have one referance that they measure in D2O but for calculation they did not put any clue. I am attaching this paper and please suggest what I should do? 

I am not entirely sure what your experimen sets out to achieve or how your experiment works exactly.  It would help if you could write down the chemical equations for the reaction/s of your experiments.

You make a good point when wondering if mass discrimination could affect the outcome of your experiments. However, to gauge this a detailed description of the chemical reactions in your study would be of great help.

I too have not understood your experience. I do not understand why you express your values in relation to the PDB. In your place I directly measure oxygen 18 relative to SMOW, on both H2O and D2O.... Also, in my opinion the scaling of your measurements between SMOW and PDB will change nothing on the interpretation of your experience...

Thank you Prof Moussa and Prof Wolfram Meier-Augenstein. The ratio of 18O/16O in unreacted natural-abundance water is designated R0. These samples were taken directly from the Millipore filtration unit. In some cases, the water was isolated by the vacuum transfer of all volatiles from solutions of 0.1 M HClO4. R0 was also determined for samples of 0.1 M HClO4 preincubated for 20 min with the reduced metal catalyst (600 μM) or CAN (0.05 M) before vacuum transfer of the liquid into glass ampules, which were then flame-sealed. A standard CO2 exchange protocol was used for analysis of H2O and consistently indicated R0 = 0.9930 ± 0.0006 versus Vienna Standard Mean Ocean Water (VSMOW). In addition, analyses were conducted at two different isotope ratio mass spectrometry (IRMS) facilities at the University of Waterloo in the Environmental Isotope Laboratory. FRom this laboratory they suppy the VPDB value both for water and D2O.  Is there any standrard R0 value for D2O.

For more information about my experiment I am attaching two related journal. Thank you for your suggession!

I do not understand your question for a standard value of R0 for (2H)2O.  Based on what you are saying, R0 of your unreacted water (be it natural abundance water of 2H labelled water) is not a constant but will depend on its particular stable isotopic composition (prior to the water oxidation experiment).

You need to understand natural abundance water does not have just one specific 18O composition.The 18O composition of terrestrial natural abundance water can range from -2 ‰ to -20 ‰.  To determine R0 you need to have your natural abundance water analyzed for its 18O composition before using it in your water oxidation experiments (which you seem to have done).  In the same way, prior to your water oxidation experiments with (2H)2O you need to have your 2H labelled water analysed for its particular 18O composition as this may (and most certainly will) be different from that of your natural abundance water. If I understand your experiments correctly, only the differences in d18O values for the reactions carried with the two different waters but under otherwise identical conditions will tell you something about potential mass discriminatory effects of the stronger 2H-O bond on your reaction.

natural abundance water experiment:  d18Odiff (1H) = d18Opre-reaction water - d18Opost-reaction water

2H labelled water experiment: d18Odiff (2H) = d18Opre-reaction 2H water - d18Opost-reaction 2H water

If d18Odiff (1H) = d18Odiff (2H) then there is no mass discriminatory effect.

If d18Odiff (1H) is significantly different from d18Odiff (2H) then there is a mass discriminatory effect.

Also, as Isaam Moussa has pointed out, for the purpose of this consideration it makes no difference if the d18O values are scaled normalized to VPDB or VSMOW as long as all d18O values are normalized to same scale.

If all waters, pre-reaction and post-reaction (for both the natural abundance and the 2H labelled water experiment) were analyzed by the same lab using the same technique (18O determination of water through equilibration with CO2 as I think they have been since results were reported on the VPDB scale) there would be no need to convert these values to the VSMOW scale since you will be comparing differences on a like-for-like basis.

If the user means that he is looking for the isotopic composition of the water molecules assumed in equilibrium with Calcite at given temperature, the above-mentioned use of the conversion equation 18O of the values of V-PDB to V-SMOW must be followed by adding the value of the isotopic separation (separation) factor (abbreviated by the Greek letter "epsilon"). In most cases the obtained d18O of such water at ambient temperature will be higher than the value of d18O in Calcite versus PDB only by 2.20 per mil. I've extensive data, diagrams and equations that exclusively show this point in details. If the user is interested in the details he may get into contact with me. He is welcome.

The Friedman equation mentioned by Dr Moussa incorporates the effect you mention (isotopic fractionation between CO2 and VSMOW water) at a set temperature of 25C°C). Incidentally, the Friedman equation has been amended and is now given (for the situation at 25°C) by the CIAAW as

d18O(VSMOW) = 1.03092×d18OVPDB + 30.92.

In section 2.3 of their article, Kim, Coplen and Horita in Geochim Cosmochim Acta (2015, vol 158, 276-289) develop this equation in full detail. Section 3 of the same paper deals with isotopic fractionation factors at 25°C and above 25°C.

I used the following equations for the data conversion:

δ18OX,VPDB = 0.97001×δ18OX,VSMOW − 29.99 ‰

δ18OX,VSMOW = 1.03092×δ18OX,VPDB +30.92 ‰

The conversion equations to convert 18O delta to VSMOW and vice versa,

δ18OVPDB = 0.97001*δ18OVSMOW - 29.99‰