Chemistry · Thermodynamics · NEET
Only the sign changes. The number stays exactly the same. Example: H2(g) + ½O2(g) → H2O(l), ΔH = −286 kJ. Reverse it: H2O(l) → H2(g) + ½O2(g), ΔH = +286 kJ. Same 286, opposite sign. Reason: ΔH is a state function. Going forward releases heat (exothermic, −), so going backward must absorb the same heat (endothermic, +).
Yes. ΔH is an extensive quantity, so it scales with the amount of substance. If C(s) + O2(g) → CO2(g) has ΔH = −393 kJ, then 2C(s) + 2O2(g) → 2CO2(g) has ΔH = 2 × (−393) = −786 kJ. Twice the moles react, so twice the heat is released.
Then you multiply ΔH by ½ too. For 2H2(g) + O2(g) → 2H2O(l), ΔH = −572 kJ. Divide the equation by 2: H2(g) + ½O2(g) → H2O(l), ΔH = −572/2 = −286 kJ. Fractional coefficients like ½O2 are allowed in thermochemical equations.
Because energy is conserved. The heat given out in the forward direction is exactly the heat you must put back in the reverse direction. The path does not matter (ΔH is a state function), so the amount is identical — only the direction of heat flow (into or out of the system) reverses, which is what the sign shows.
Do them in order. First reverse: flip the sign only. Then multiply: scale the (already flipped) number by the factor. Example: reverse and double CH4 combustion (ΔH = −890 kJ). Step 1 reverse → +890 kJ. Step 2 multiply by 2 → +1780 kJ. This is exactly how Hess's Law problems combine equations.
In Hess's Law of Constant Heat Summation questions. You are given a few thermochemical equations and a target equation. You reverse some (flip signs) and multiply some (scale ΔH) so that when you add them, they give the target reaction. Then you add the adjusted ΔH values to get the answer. These two rules are the whole engine of Hess's Law arithmetic.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. Reversing only flips the sign (+ to − or − to +). The magnitude (the number) stays exactly the same because ΔH is a state function.
ΔH is multiplied by 3 as well. ΔH is an extensive property, so it scales directly with the coefficients of the balanced equation.
Yes. Because ΔH is quoted per the amounts shown, fractions are allowed, for example H2(g) + ½O2(g) → H2O(l). If you clear the fraction by doubling, you must double ΔH too.
These two rules power Hess's Law questions. You reverse and multiply given equations so they add up to the target reaction, then add the adjusted ΔH values to find the unknown enthalpy.
The states (s, l, g) stay attached to the same species; you only swap which side they are on. Products become reactants and vice versa, but H2O(l) stays H2O(l).
