Calculate the enthalpy change for the reaction:

2S(s) + 3O2(g) --> 2SO3(g)

Use the following thermochemical equations:

Reaction
S(s) + O2(g) --> SO2(g) Delta H(in kJ) -296.8
2SO2(g) + O2(g) --> 2SO3(g) Delta H(in kJ) -197.0

To calculate the enthalpy change for the reaction, we can use the given thermochemical equations and apply Hess's Law.

The given reaction consists of two steps:
1) S(s) + O2(g) --> SO2(g) with ΔH = -296.8 kJ
2) 2SO2(g) + O2(g) --> 2SO3(g) with ΔH = -197.0 kJ

We need to flip the first equation and multiply the second equation by 2 so that the reactants and products match the desired reaction:
1) SO2(g) --> S(s) + O2(g) with ΔH = +296.8 kJ
2) 4SO2(g) + 2O2(g) --> 4SO3(g) with ΔH = -394.0 kJ

Now, sum up the two equations to obtain the desired reaction:
4SO2(g) + 2O2(g) --> 4SO3(g) with ΔH = -394.0 kJ
-2S(s) + 3O2(g) --> 2SO3(g) with ΔH = -394.0 kJ

Therefore, the enthalpy change for the reaction 2S(s) + 3O2(g) --> 2SO3(g) is -394.0 kJ.

To calculate the enthalpy change for the given reaction, we can use Hess's law. According to Hess's law, the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps or reactions that make up the overall reaction.

In this case, we have the following thermochemical equations:

1. S(s) + O2(g) --> SO2(g) ΔH = -296.8 kJ
2. 2SO2(g) + O2(g) --> 2SO3(g) ΔH = -197.0 kJ

We want to find the enthalpy change for the reaction: 2S(s) + 3O2(g) --> 2SO3(g)

To use these equations and obtain the desired equation, we need to manipulate them to match the desired reaction. We can multiply the first equation by 2 to get a balanced equation with 2SO2:

2S(s) + 2O2(g) --> 2SO2(g) ΔH = -593.6 kJ

Now, we can manipulate the equations to get the desired reaction:

2S(s) + 2O2(g) --> 2SO2(g) ΔH = -593.6 kJ
2SO2(g) + O2(g) --> 2SO3(g) ΔH = -197.0 kJ

By adding these two equations together, we can cancel out the intermediate compound SO2 and obtain the desired reaction:

2S(s) + 3O2(g) --> 2SO3(g) ΔH = -593.6 kJ + (-197.0 kJ) = -790.6 kJ

Therefore, the enthalpy change for the given reaction is -790.6 kJ.

2NH2+ H2=2NH3...delta H=?

Multiply equation 1 by 2 (multiply delta H by 2 also), add to equation 2 as is. Add deltaHs for total.