Reactions subtracting

Take the theoretical yield of the product and determine how much of the reactant in excess is used up in the reaction. Subtract that from the starting amount to find the amount left. 

Both have the same steric and conjugative effects, but the polar effect is largely absent in the acid-catalyzed reactions. Subtraction of the logarithms of the two relative rates therefore leaves just the polar effect. This is written... 

Equations 12, 14 and 17 require the presence of H20. Thus H20 plays an important role in promoting the catalytic activity, but can also cause deactivation. Catalysis will be more efficient when all the reactions directly involved in the catalytic cycle are faster than the side reactions subtracting active species. Deactivation is related to the requirement of the palladium centre to have a vacant coordination site to ensure high catalytic activity. However, palladium tends to achieve the usual coordination number four, for example through dimerisation. Dimerisation/deactivation can be prevented by coordination of labile ligands, like H20, which acts also as an efficient hydride source. Also deprotonation leads to dimerisation/deactivation an acid can prevent it. 

Write the two half-reactions for the following redox reaction. Subtract the two reduction potentials to find the standard cell potential for a galvanic cell in which this reaction occurs. 

Step 5 To write a balanced net cell reaction, subtract the left half-reaction from the right halfreaction. (Subtraction is equivalent to reversing the left-half reaction and adding.)... 

To calculate A S° for the reaction, subtract the standard molar entropies of all the reactants from the standard molar entropies of all the products. Look up the S° values in Table 17.1 or Appendix B and remember to multiply the S° value for each substance by its coefficient in the balanced chemical equation. 

All the magnesium should be used up by the reaction, since it is the limiting reagent. If any magnesium ribbon does remain after the reaction, rinse it with water, dry it with a paper towel, and measure its mass. To find the mass of the magnesium used up by the reaction, subtract the final mass from the initial mass. 

Interpreting Data What mass of strontium sulfate is produced in the reaction Subtract the mass of filter paper from the mass of strontimn sulfate plus filter paper. 

To calculate the enthalpy change of a chemical reaction, subtract the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products ... 

Step 3. Subtract the second reaction from the first this regenerates the original reaction. Subtracting the second potential from the first yields = 0.771 — 0.15 = 0.62 V. 

Criegee glycol cleavage reaction Subtractive oxidation... 

Strategy To express a reaction as the difference of two reduction halfreactions, identify one reactant species that undergoes reduction and its corresponding reduction product, then write the half-reaction for this process. To find the second half-reaction, subtract the first half-reaction from... 

The creatinine formed can be measured by the Jaffe reaction. Subtraction of the creatinine value of the untreated sample gives the creatine level. 

If we compare the calculated total ionization potential, IP = 4.00 hartiees, with the experimental value, IP = 2.904 hartiees, the result is quite poor. The magnitude of the disaster is even more obvious if we subtract the known second ionization potential, IP2 = 2.00, from the total IP to find t c first ionization potential, IPi. The calculated value of IP2, the second step in reaction (8-21) is IP2 = Z /2 = 2.00, which is an exact result because the second ionization is a one-election problem. For the first step in reaction (8-21), IPi (calculated) = 2.00 and IPi(experimental) = 2.904 — 2.000 =. 904 hartiees, so the calculation is more than 100% in error. Clearly, we cannot ignore interelectronic repulsion. 

To a first approximation, the activation energy can be obtained by subtracting the energies of the reactants and transition structure. The hard-sphere theory gives an intuitive description of reaction mechanisms however, the predicted rate constants are quite poor for many reactions. 

Subtracting reaction 2 from reaction 1 gives the familiar water gas shift reaction (eq. 3). 

The small value of the entropy change reflects the fact that only liquids are involved in tlris reaction. The heat balance in canying out tlris reaction may be calculted, according to Hess s law, by calculating tire heat change at room temperarnre, and subtracting tire heat required to raise the products to the hnal teirrperamre. The data for tlris reaction are as follows ... 

But this is not the whole story We not only need to know that a trajectory that crosses the transition state surface is eventually deactivated as product, we also need to know whether it originated from the reactant well A trajectory that originates from the product well and ends up as product won t contribute to the forward rate of reaction. Some of the trajectories did originate as product. We need to find that fraction and subtract it. 

As has been described earlier, the stoichiometric reactions should be manipuitUed algebraically to retain the transferable species (H2S) only in the first equation. Therefore, HjS can be eliminated from Eq. (8.13) by subtracting (8.12) from (8.13) to get... 

X 10 14Af (by a factor of 107), reaction (55) must consume about 10-7 mole of hydroxide ion for every liter of solution. Since one mole of OH-(ag) reacts with one mole of H+(aq), the amount of H+(aq) required is also 10-7 mole for every liter of solution. Subtracting 10-7 mole/liter from a concentration near 1 mole/liter causes such a small change in [H+] that it need not be considered in calculations (such as in Exercises 11-1 and 11-2). 

---