Given the following data use Hesss Law to compute the enthal

Given the following data, use Hess\'s Law to compute the enthalpy change for the combustion of acetylene, C2H2(8). (You\'ll first have to write the balanced combustion equation for acetylene) H2(g) + ½ O2(g) H2O(l) 9. Ho =-285.5 kJ Ho =-393.5 kJ C(graphite) +02(g) CO2(g) 2C(graphite) + H2(g) C2H2(g) Ho= 226.7 kJ 10. Calculate the heat of formation of Mn02(s) (i.e. the enthalpy change when MnO2(s) is formed from its elements: Mn(s) + O2(g) MnO2(s)) by using Hess\'s law and the following reaction data 2 MnO(s) 2 Mn(s) + O2(g) H= 2Mnols) + 02(g) 2 MnO2(s) 770.4k/ H=-269.7 ki 3

Solution

Solution:- (9) The equation for the combustion of acetylene is as follows...

2C₂H₂(g) + 5O₂(g) ------> 4CO₂(g) + 2H₂O(l) delta H = ?

Now, we would make this equation from the given equations.

First of all we want two C₂H₂(g) which is present in third equation but on product side. So, we need to reverse this equation and also need to multiply it by 2.

2C₂H₂(g) -----> 4C(graphite) + 2H₂(g) delta H = -453.4 kJ

next we need O2 but since it is present in more than one equations, we would take a look at this later.

Next is CO₂(g) and it is present in second equation but we need to multiply the second equation by 4.

4C(graphite) + 4O₂(g) ----> 4CO₂(g) delta H = -1574 kJ

Next we need two H₂O and it is present in first equation, We need to multiply the equation by 2.

2H₂(g) + O₂(g) ----> 2H₂O(l) delta H = -571 kJ

On adding all these three equations we get our desired equation,

2C₂H₂(g) -----> 4C(graphite) + 2H₂(g) delta H = -453.4 kJ

4C(graphite) + 4O₂(g) ----> 4CO₂(g) delta H = -1574 kJ

2H₂(g) + O₂(g) ----> 2H₂O(l) delta H = -571 kJ

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2C₂H₂(g) + 5O₂(g) ------> 4CO₂(g) + 2H₂O(l) delta H = -2598.4 kJ

(10) We need to make the equation,----

Mn(s) + O₂(g) -------> MnO₂(s) delta H = ?

Mn(s) is present in first equation so need to reverse it and also divide by 2.

Mn(s) + 1/2O₂(g) ----> MnO(s) delta H = -385.2 kJ

Now, we need to divide the second equation by 2 as it has two MnO₂(s) where as we beed only one.

MnO(s) + 1/2 O₂(g) ----> MnO₂(s) delta H = -134.85 kJ

On adding these two equations...

Mn(s) + 1/2O₂(g) ----> MnO(s) delta H = -385.2 kJ

MnO(s) + 1/2 O₂(g) ----> MnO₂(s) delta H = -134.85 kJ

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Mn(s) + O₂(g) ----------> MnO₂(s) delta H = -520.1 kJ