[10:58am]: Wn 4.5g of liquid water was formed by burning hydrogen gas in oxygen,-72kj of heat was given off.calc d standad heat formation of water
2H2 + O2 ==> 2H2O + heat
So 4.5 g gives 72 kJ, heat for the reaction above is
-72 kJ x 36 g/4.5 g = -576 kJ. That's for 2 mols so for 1 mole is it -576/2 = ?
Well, well, well, looks like we have some chemistry on our hands! Don't worry, I'll help you out in my own hilarious way!
To determine the standard heat formation of water, we need to use a balanced chemical equation and some stoichiometry magic.
First, let's start with the balanced equation for the combustion reaction of hydrogen gas (H2) and oxygen gas (O2) forming water (H2O):
2H2 + O2 → 2H2O
Now, we know that 4.5g of water was formed and -72 kJ of heat was given off. We need to find the standard heat formation, so let's get to it!
Step 1: Convert the mass of water to moles: 4.5g ÷ molar mass of water (18g/mol) = 0.25 moles of water.
Step 2: Remember that the balanced equation tells us that for every 2 moles of water formed, we get -72 kJ of heat released. So, if we have 0.25 moles of water, the heat released would be (-72 kJ ÷ 2) × 0.25 = -9 kJ.
Voila! The standard heat formation of water is -9 kJ. Keep in mind that standard heat formation values are usually reported per mole, so don't forget about that!
Hope this little chemistry comedy routine helped you out!
-288kJ
To calculate the standard heat of formation of water (ΔHf°), we need to use the equation:
ΔHf° = ΣΔHf°(products) - ΣΔHf°(reactants)
First, we need to identify the reactants and products in the given reaction. From the information provided, we know that the reactants are hydrogen gas (H2) and oxygen gas (O2), and the product is liquid water (H2O).
Now, we need to find the values of the standard heat of formation (ΔHf°) for each compound involved. The standard heat of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
The standard heat of formation values for H2(g), O2(g), and H2O(l) are as follows:
ΔHf°(H2) = 0 kJ/mol (by definition)
ΔHf°(O2) = 0 kJ/mol (by definition)
ΔHf°(H2O) = -285.8 kJ/mol
Now, substitute these values into the equation:
ΔHf° = ΣΔHf°(products) - ΣΔHf°(reactants)
ΔHf° = [ΔHf°(H2O)] - [ΔHf°(H2) + ΔHf°(O2)]
Plugging in the values:
ΔHf° = [-285.8 kJ/mol] - [0 kJ/mol + 0 kJ/mol]
ΔHf° = -285.8 kJ/mol
Therefore, the standard heat of formation of water is -285.8 kJ/mol.
To determine the standard heat formation of water (ΔHf°), we need to use the equation:
ΔHf° = ΣΔHf°(products) - ΣΔHf°(reactants)
Where:
- ΣΔHf°(products) is the sum of the standard heat formation values for the products
- ΣΔHf°(reactants) is the sum of the standard heat formation values for the reactants
In this case, we have the balanced chemical equation for the reaction:
2H2(g) + O2(g) → 2H2O(l)
We are given that 4.5 grams of liquid water (H2O) is formed and that -72 kJ of heat was given off. However, we need to convert the given mass of water into moles before we can proceed.
To convert grams to moles, we need to use the molar mass of water, which is approximately 18 g/mol. Thus:
moles of water = mass of water / molar mass of water
= 4.5 g / 18 g/mol
= 0.25 mol
Now, we have the moles of water formed in the reaction. Looking at the balanced equation, we can see that 2 moles of water are formed for every 2 moles of hydrogen gas (H2) consumed. Therefore, the moles of hydrogen gas consumed can be calculated as:
moles of H2 = (moles of water) / 2
= 0.25 mol / 2
= 0.125 mol
Since the reaction is balanced, the moles of hydrogen gas consumed will be the same as the moles of oxygen gas consumed. So, we also have 0.125 mol of oxygen gas (O2) consumed.
Now we can calculate the standard heat formation of water:
ΔHf° = ΣΔHf°(products) - ΣΔHf°(reactants)
For each component, we need to look up the standard heat formation values in a reference table. The standard heat formation of hydrogen gas (H2) and oxygen gas (O2) is zero since they are in their standard states.
Given that the standard heat formation of water is -285.8 kJ/mol, we can calculate:
ΔHf° = (2 mol H2O) * (-285.8 kJ/mol) - [(0.125 mol H2) * (0 kJ/mol) + (0.125 mol O2) * (0 kJ/mol)]
= -571.6 kJ - 0 kJ
= -571.6 kJ
Therefore, the standard heat formation of water (ΔHf°) is -571.6 kJ.