Calculate the standard heat of formation of acetaldehyde , 2C(s) + 2 H2(g) +1/2 O2 (g) ---> CH3CHO(g) given the following information?

CH3CHO(g) + 5/2 O2 (g) --> 2H2O(l) + 2 CO2(g) DeltaH rxn =-1194 KJ

H2(g)+ 1/2 O2 (g) --> H2O (l) DeltaH rxn =-286 KJ/mol

C(s) + O2(g) ---> CO2(g) DeltaH rxn=-394 KJ/mol

Ive tried flipping first equation then multiplying second equation and 3rd equation but I cannot get the correct answer which is -170 Kj . plus I can't get oxygen to 1/2 . PLEASE HELP ME AND SHOW ME WORK! THANKS !!

Reverse equation 1 and add to twice eqn 2 and twice equn 3. I didn't add dH to see that it comes out to -170 kJ but the equation you want is right so I assume the numbers will work.

To calculate the standard heat of formation of acetaldehyde (CH3CHO), we need to use the given information about the standard enthalpies of formation of the reactants and products involved in the reaction. The standard enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states.

First, let's write the balanced equation for the formation of acetaldehyde (CH3CHO):

2C(s) + 2 H2(g) + 1/2 O2(g) -> CH3CHO(g)

Now, we can use the given information to calculate the standard heat of formation.

Step 1: Flip the first equation and change the sign of its enthalpy value:

C(s) + O2(g) -> CO2(g) ΔH = -(-394 KJ/mol) = +394 KJ/mol

Step 2: Multiply the second equation by 2 (since we need 2 moles of H2O):

2[H2(g) + 1/2 O2(g) -> H2O(l)] ΔH = 2*(-286 KJ/mol) = -572 KJ/mol

Step 3: Multiply the flipped equation (from Step 1) by 2 (since we need 2 moles of CO2):

2[C(s) + O2(g) -> CO2(g)] ΔH = 2*(+394 KJ/mol) = +788 KJ/mol

Step 4: Add the equations together to cancel out common compounds:

2[H2(g) + 1/2 O2(g) -> H2O(l)] + 2[C(s) + O2(g) -> CO2(g)] = -572 KJ/mol + 788 KJ/mol

This yields the equation:

2H2(g) + 1/2 O2(g) -> 2H2O(l) + 2CO2(g) ΔH = +216 KJ/mol

Step 5: Finally, subtract the equation obtained in Step 4 from the given equation to obtain the standard heat of formation of acetaldehyde:

CH3CHO(g) + 5/2 O2(g) - (2H2O(l) + 2CO2(g)) = -1194 KJ/mol - (+216 KJ/mol)

Simplifying the equation and subtracting the values:

CH3CHO(g) + 5/2 O2(g) -> -1410 KJ/mol

So, the standard heat of formation of acetaldehyde (CH3CHO) is -1410 KJ/mol.

Please note that the value obtained here (-1410 KJ/mol) doesn't match the given correct answer (-170 KJ). There might be a mistake in the given information or the calculations. Double-check the given values and the calculations to ensure accuracy.