use standard enthalpies of formation to calculate delta Hrxn for the following reaction. 2H2S(g)+3O2(g)-->2H2O(l)+2SO2(g) express the answer using four significant figures

Delta Hrxn = ? kJ

ÄH°reaction = sum of standard enthalpy of formation of products - sum of standard enthalpy of formation of reactants

ÄH°reaction = [ 2 X ÄH°f for H2O + 2 X ÄH°f for SO2 ] - [ 2 X ÄH°f for H2S + 3 X ÄH°f for O2 ]
ÄH°reaction = [ 2 X -285.8 + 2 X -296.8 ] - [ 2 X -20.63 + 3 X 0 ]
ÄH°reaction = [-571.6 - 593.6 ] - (-41.26)
ÄH°reaction = -1165.2 + 41.26 = -1123.94 kj

Well, I'm not a scientist, but I'll give it a shot! And by shot, I mean a humorous response. So, let's dive into it and see what we get!

To calculate the enthalpy change for this reaction, we need to look up the standard enthalpies of formation for each compound involved. It's like checking a recipe to see how much spice you need to add to make it sizzle!

Alright, here we go:

The standard enthalpy of formation for H2S(g) is -20.6 kJ/mol.
For O2(g), it's 0 kJ/mol.
For H2O(l), it's -285.8 kJ/mol.
And for SO2(g), it's -296.8 kJ/mol.

Now, let's shake things up and calculate the enthalpy change:

Delta Hrxn = (2 * -285.8 kJ/mol) + (2 * -296.8 kJ/mol) - (2 * -20.6 kJ/mol) - (3 * 0 kJ/mol)

After summing up those values, my calculation gives me:

Delta Hrxn = -1175.8 kJ

Voila! The enthalpy change for the reaction is approximately -1175.8 kJ. Remember, I'm just a clown bot, so don't take it too seriously.

To calculate ΔHrxn for the given reaction using standard enthalpies of formation, we need to use the equation:

ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)

First, we need to identify the standard enthalpies of formation (ΔHf) for each of the compounds involved in the reaction. Let's look these up:

ΔHf(H2S) = -20.6 kJ/mol (from a reference table)
ΔHf(O2) = 0 kJ/mol (Oxygen is in its standard state)
ΔHf(H2O) = -285.8 kJ/mol (from a reference table)
ΔHf(SO2) = -296.8 kJ/mol (from a reference table)

Now, we can substitute these values into the equation for ΔHrxn:

ΔHrxn = (2 mol * ΔHf(H2O)) + (2 mol * ΔHf(SO2)) - (2 mol * ΔHf(H2S)) - (3 mol * ΔHf(O2))

ΔHrxn = (2 * -285.8 kJ/mol) + (2 * -296.8 kJ/mol) - (2 * -20.6 kJ/mol) - (3 * 0 kJ/mol)

Simplifying:

ΔHrxn = -571.6 kJ/mol - 593.6 kJ/mol + 41.2 kJ/mol

ΔHrxn = -1124 kJ/mol

Now, we need to express the answer using four significant figures:

ΔHrxn = -1120 kJ/mol

Therefore, the ΔHrxn for the given reaction is -1120 kJ/mol.

To calculate the standard enthalpy change (ΔHrxn) for a reaction using standard enthalpies of formation, you need to follow these steps:

Step 1: Identify the standard enthalpies of formation (∆Hf) for each compound involved in the reaction. Standard enthalpies of formation represent the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure.

Here are the standard enthalpies of formation (∆Hf) for the compounds:

ΔHf[H2S(g)] = -20.2 kJ/mol
ΔHf[O2(g)] = 0 kJ/mol
ΔHf[H2O(l)] = -285.8 kJ/mol
ΔHf[SO2(g)] = -296.8 kJ/mol

Step 2: Write the balanced chemical equation for the reaction:

2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)

Step 3: Calculate the ΔHrxn using the stoichiometric coefficients and the standard enthalpies of formation. The ΔHrxn is the sum of the products' ΔHf values minus the sum of the reactants' ΔHf values, multiplied by their respective coefficients:

ΔHrxn = [2ΔHf[H2O(l)] + 2ΔHf[SO2(g)]] - [2ΔHf[H2S(g)] + 3ΔHf[O2(g)]]

Plugging in the values:

ΔHrxn = [2(-285.8 kJ/mol) + 2(-296.8 kJ/mol)] - [2(-20.2 kJ/mol) + 3(0 kJ/mol)]

Calculating:

ΔHrxn = [-571.6 kJ/mol - 593.6 kJ/mol] - [-40.4 kJ/mol]

ΔHrxn = [-1165.2 kJ/mol] + [40.4 kJ/mol]

ΔHrxn = -1124.8 kJ/mol

Therefore, the ΔHrxn for the given reaction is -1124.8 kJ/mol.