Question

In order to decompose 9 grams of water 142.5 kJ heat is required. Hence the enthalpy of formation of water is:
(a) +142.5 kJ
(b) -142.5 kJ
(c) +285 kJ
(d) -285 kJ

Answer 1

Hint: In the decomposition of water, the products are hydrogen and oxygen and in the formation of water hydrogen and oxygen combine to form water. Calculate the enthalpy of the reaction with the number of moles and when the reaction is reversed then the sign of the enthalpy also changes.

Complete step by step solution:
In decomposition of 1 mole of water there is formation of 1 mole of hydrogen and half mole of oxygen and the reaction is given below:
${{H}_{2}}O\to {{H}_{2}}+\dfrac{1}{2}{{O}_{2}}$
So, we are given the information that for the decomposition of 9 grams of water, 142.5 kJ of heat is required. Therefore, the number of moles of water can be calculated:
$Moles=\dfrac{given\text{ }mass}{molecular\text{ }mass}$
The molecular mass of water (${{H}_{2}}O$) will be: 1 (2) + 16 = 2 + 16 = 18
$Moles=\dfrac{9}{18}=\dfrac{1}{2}$

So, the 142.5 kJ of heat is required to decompose $\dfrac{1}{2}$ mole of water, but in the reaction one mole of water is there. Therefore, for the decomposition of one mole of water, the heat will be:
$142.5\text{ x 2 = 285 kJ}$
Therefore, in the reaction we can write:
${{H}_{2}}O\to {{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\text{ ; }\Delta \text{H = 285 kJ}$

Now, the reaction of formation of water, the reactants is hydrogen and oxygen and the reaction will be:
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$
So, we can see that this reaction is the reverse reaction of decomposition, so the enthalpy of this reaction will be negative, this is given below:
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O\text{ ; }\Delta \text{H = -285 kJ}$
So, the correct answer is “Option D”.

Note: You can change the sign of the enthalpy of the reaction only if the number of moles of all the compounds in the reaction is the same, if the number of moles changes then the enthalpy of the reaction will also change accordingly.