Choose the thermochemical reaction below that represents...

Choose the thermochemical reaction below that represents $\Delta H^o_f$. 2 Li(s) + F$_2$(g) → 2 LiF(s) Sr(g) + Br$_2$(g) → SrBr$_2$(s) 2 Na(s) + C(diamond) + $\frac{3}{2}$ O$_2$(g) → Na$_2$CO$_3$(g) Mg(s) + C(graphite) + $\frac{3}{2}$ O$_2$(g) → MgCO$_3$(s) Question 10 (2 points) Saved

Transcript

00:01 Hi now we will discuss about delta h not of the reaction and delta h0 f of the products we know that delta h not of reaction is equal to it is the sum of the delta h0 of the products minus sum of the delta h0 values of the reactions in a chemical reaction and delta h0 f is the change in enthalpy of the the formation and this is the change in enthalpy of the formation for the formation of one more of substance during the formation of one mole of substance and for the formation of one more of substance the constituent elements should be in their standard state elements in their standard state under standard conditions so this is said to be delta h -0 the change in enthalpy of formation for of formation of one mole of substance when the elements are in their standard states under standard conditions is set to be the change in enthalpy of formation now we have provided with a few chemical equations in those reactions which reactions are processing delta h0 reaction is equal to delta h0 f of the product we have to check we have provided with the following equations two n a s plus f2 gas gives rise to 2n a f solid we write the delta h not of the reaction for this so it is written as delta h0 formation of n a f as it is two molecules should be multiplied with 2 minus reactants reactants are sodium and fluorine delta h0 formation of this is also 2 times should be multiplied with 2 plus delta h0 f of f2 here delta h not reaction is not equal to the delta h0 f of n a because here one mole of substance should form here two moles of substance is forming so it is incorrected in this reaction delta h0 reaction delta h0 reaction is not equal to delta h0 of product now we will see the other reaction baco3 solid use rise to b ao solid plus co2 gas for this if we write the delta h of the reaction delta h not of the reaction is equal to delta h not formation of b ao plus delta h h not formation of p .o2 minus delta h0 formation of p a c o 3 here it is not a formation reaction here the decomposition reaction so delta h0 reaction never equals to the delta h0 so for this reaction delta h0 reaction is not equal to delta h0 formation of the product now we will write the delta h0 reaction for the given reaction n a plus half f2 gives rise to n a f f f f f f f f f f f f f so here delta h not formation of n a f minus delta h not formation of n a plus delta h not formation of half f2 so here f2 is in its liquid state so delta h0 reaction never equal to the delta h0 reaction never equals to the delta h0 f of product if we write the delta h not of this reaction delta h0 f of c o minus delta h h not f of c o plus delta h not f of half foot so here c o is not an element so here also the delta h0 of the reaction is never equal to the delta h0 f of products if we write the delta h0 of the reaction the solid graphed plus o2 gives rise to c o2 gas can be written as delta h not formation of c o2 minus delta h0 formation of d plus delta h0 formation of here see in graphite it is in its standard state so its value is 0 and delta h0 formation of o2 is in its gaseous state it's in standard state so this value also zero so here the delta h0 reaction is equal to delta h0 f of product now we will see another equation if we write the delta h0 of this reaction, nia solid plus half f2 gives rise to nif solid.

06:22 Delta h0 of the reaction is equal to delta h0 formation of naf minus delta h0 f of sodium plus delta h0 f of half...

Question

Choose the thermochemical reaction below that represents $\Delta H^o_f$. 2 Li(s) + F$_2$(g) → 2 LiF(s) Sr(g) + Br$_2$(g) → SrBr$_2$(s) 2 Na(s) + C(diamond) + $\frac{3}{2}$ O$_2$(g) → Na$_2$CO$_3$(g) Mg(s) + C(graphite) + $\frac{3}{2}$ O$_2$(g) → MgCO$_3$(s) Question 10 (2 points) Saved

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