For the reaction 2H_(2)(g)+O_(2)(g)arrow 2H_(2)O(g) the bond enthalpies (inkJ/mol) are H-H=x,O=O=y,O-H=z Which calculation will give the value, in kJ/mol, of Delta H for the reaction? 4z-2x-y 2z-2x-y 2x+y-4z 2x+y-2z

To determine the value of $\Delta H$ for the reaction $2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)$, we need to consider the bond enthalpies of the bonds broken and formed during the reaction.<br /><br />In the reaction, the bonds broken are:<br />- 2 H-H bonds in $2H_{2}(g)$<br />- 1 O=O bond in $O_{2}(g)$<br /><br />The bonds formed are:<br />- 4 O-H bonds in $2H_{2}O(g)$<br /><br />The calculation for $\Delta H$ is:<br />$\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}$<br />$\Delta H = (2x + y) - 4z$<br /><br />Therefore, the correct calculation is $2x+y-4z$.