- Calculate ΔH^0 values at 25^∘ C for the reactions of 1 mol of each of the following metals wi

- Calculate ΔH^0 values at 25^∘ C for the reactions of 1...

- Calculate $\Delta H^{0}$ values at $25^{\circ} \mathrm{C}$ for the reactions of $1 \mathrm{~mol}$ of each of the following metals with stoichiometric quantities of water to form metal hydroxides and hydrogen: (a) $\mathrm{Li},(\mathrm{b}) \mathrm{K},$ and (c) Ca. Rationalize the differences in these values.

Transcript

00:01 In answering this question, let's first of all write the balanced chemical equation where we have lithium reacting with h2o to produce lithium hydroxide and half a mole of hydrogen.

00:13 Now the standard enthalpy change of a reaction is equal to the sum of the entropy changes of formation of the products minus the sum of the entropy changes of formation of the reactants under standard conditions.

00:30 Looking at this reaction, what we have is we are looking at the sum of the endopi changes of formation of the products, which is lithium and half a mole of hydrogen.

00:42 So this is going to be 1 multiplied by negative 487 .23 kilojoules per more plus half multiplied by 0 kilochols per more, which is the standard enthalpy change of formation of hydrogen.

00:58 And then we subtract the enthalpy change of formation of the lithium and the h2o.

01:05 And this is going to be equal to negative 1 multiplied by 0, which is that of lithium.

01:17 And to this we say minus 285, 285 .8 kilojoules per mole that of h2o.

01:26 This gives us standard enthalpy change of reaction of negative to 0 .4 in 3805 .4.

01:33 Kilojoules per and then moving on to the next reaction if we apply a similar approach we have potassium reacting with h2o to produce potassium hydroxide and half h2 so if we apply the similar approach to that above we are going to get negative 138 .9 in kilojoules per more and then for the next reaction where we have calcium we are looking at calcium and h2o reacting to form calcium hydroxide and hydrogen so here we've got two moles of h2o so the enthalpy change of this reaction under standard conditions this is going to be the end up the change of formation of calcium which is negative 986 kilojoules per more plus that of hydrogen which is equal to zero minus that of two moles of h2o, which is negative 28, 285, negative 285, minus that of calcium, which is 0, giving us a standard enthalpy change of reaction that is equal to negative 4, 15 in kilojoules per more.

03:04 So, in explanation, we can look at the enthalpy changes and compare them against the electrode potentials.

03:14 So if we are looking at lithium ion, potassium ion, and we look at calcium ions, the electrode potential.

03:23 Under standard conditions here, we've got negative 3 .045...

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