Question

1.
When H₂(g) reacts with O₂(g) to form H₂O(g) , 242 kJ of energy are evolved for each mole of H₂(g) that reacts.

Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive.

Use the SMALLEST INTEGER coefficients possible and put the energy term (including the units) in the last box on the appropriate side of the equation. If a box is not needed, leave it blank.

2.
A scientist measures the standard enthalpy change for the following reaction to be -2919.0 kJ:

2C₂H₆(g) + 7 O₂(g)4CO₂(g) + 6 H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(g) is  kJ/mol.

3.
When Na(s) reacts with H₂O(l) to form NaOH(aq) and H₂(g) , 184 kJ of energy are evolved for each mole of Na(s) that reacts.

Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation. Note that the answer box for the energy term is case sensitive.

Use the SMALLEST INTEGER coefficients possible and put the energy term (including the units) in the last box on the appropriate side of the equation. If a box is not needed, leave it blank.

Answer 1

1.   H₂(g) + 1/2 O₂(g) ----------> H₂O(g) + 242 kJ (as the heat is evolved so positive sign is given)

We can also write like ; 2 H₂(g) + O₂(g) ---------->  2 H₂O(g) + 484 kJ (multiplying by 2)

2. 2C₂H₆(g) + 7 O₂(g) -------------->  4CO₂(g) + 6 H₂O(g) ; Enthalpy change= -2919.0 kJ

Standard Enthalpy of formation of C₂H₆ = -83.8 Kj/mole

Standard Enthalpy of formation of  O₂ = 0 KJ/mole

Standard Enthalpy of formation of CO₂ = -393.5

( you can find these data in any std. book or simply google it )

From the above reaction we can write as;

6* Standard Enthalpy of formation of H₂O + 4* Std. Enthalpy of formation of CO₂ - 2*Standard Enthalpy of formation of C₂H₆ - 7*Standard Enthalpy of formation of  O₂ = -2919 Kj

=> 6*Standard Enthalpy of formation of H₂O + 4*(-393.5 KJ/mol) - 2*(-83.8 Kj/mole) - 7*(0) = -2919 KJ/mol

=> Standard Enthalpy of formation of H₂O = 290.26 KJ/mol

3. Na(s) + H₂O(l) = NaOH(aq) + H₂(g) + 184 KJ/mole of Na

On balancing, we get; 2 Na(s) + 2 H₂O(l) = 2 NaOH(aq) + H₂(g) + 368 Kj (as 2 moles of Na are involved)

Please give it a thumbsup if you like my answer. Feel free to ask your queries in the comment box. Thank you