3 deleted 10 characters in body edited Jul 17 at 14:53 Buck Thorn 6,06322 gold badges88 silver badges3434 bronze badges The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$$H$$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (moremore negative) for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $$H_A>H_B$$. The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$$H$$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (more negative) for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $$H_A>H_B$$. The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$$H$$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is more negative for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $$H_A>H_B$$. 2 added 121 characters in body edited Jul 17 at 14:45 Buck Thorn 6,06322 gold badges88 silver badges3434 bronze badges The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, referring to stating "$$H$$ for B is less than for A as less itA" might not bemake it clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (more negative) for A than for B. Note that where we place A and B is not important herefor the purposes of this illustration, only that Awhere we compare the relative enthalpies qualitatively (not quantitatively). What's important is above Bthat $$H_A>H_B$$. The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, referring to $$H$$ for B than for A as less it might not be clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (more negative) for A than for B. Note that where we place A and B is not important here, only that A is above B. The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, stating "$$H$$ for B is less than for A" might not make it clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (more negative) for A than for B. Note that where we place A and B is not important for the purposes of this illustration, where we compare the relative enthalpies qualitatively (not quantitatively). What's important is that $$H_A>H_B$$. 1 answered Jul 16 at 19:21 Buck Thorn 6,06322 gold badges88 silver badges3434 bronze badges The following schematic (not meant to represent the actual changes in $$H$$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "less" when comparing signed quantities. For instance, referring to $$H$$ for B than for A as less it might not be clear whether one means "more negative" or "smaller in magnitude". In this schematic it is clear that $$\Delta \Delta _\text{f} H(A-B)>0$$, that is, A has greater (more positive) enthalpy of formation than B. On the other hand $$\Delta \Delta _{\text{comb}} H(A-B)<0$$, that is, the heat of combustion is greater (more negative) for A than for B. Note that where we place A and B is not important here, only that A is above B.