Question

Calculate the ΔH for the reaction 4 CO2 (g) + 2 H2O (g) → 2 C2H2 (g) +5 O2 (g), from the following:C2H2 (g) + 2 H2 (g) → C2H6 (g)ΔH = – 94.5 kJH2O (g) → H2 (g) + ½ O2 (g)ΔH = + 241.8 kJC2H6 (g) + 3 ½ O2 (g) → 2 CO2 (g) + 3 H2O (g)ΔH = – 283 kJ

Answer 1

Step-by-step explanation:

Here, we want to calculate the change in enthalpy of the given reaction

To get that, we are going to use the 3 related equations below it

We proceed as follows:

`\begin{gathered} We\text{ multiply equation ii by 3} \\ The\text{ heat becomes } \\ 3\text{ }*\text{ 241.8 = +725.4 KJ} \end{gathered}`

We arrange equation 1 in the opposite direction

This makes the heat become:

`+94.5\text{ KJ}`

For equation iii, we arrange in opposite direction too

The heat becomes:

`-725.4\text{ KJ}`

We can now begin to make additions and subtractions as follows: