Consider the following reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) ÄH = -891 kJ
Calculate the enthalpy change for each of the following cases.
(a) 5.00 g methane is burned in excess oxygen.
You omitted the arrow. We can't separate products from reactants without an arrow.
-891 kJ x (5.00/16) = ?? kJ.
To calculate the enthalpy change for the given reaction, we need to use the given enthalpy change value (-891 kJ) and the stoichiometric coefficients of the reactants and products.
First, let's convert the given mass of methane (CH4) into moles:
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.01 g/mol) (4H) = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol
Moles of CH4 = mass of CH4 / molar mass of CH4 = 5.00 g / 16.05 g/mol = 0.311 mol
Since the reaction equation shows that 1 mole of CH4 reacts to produce 891 kJ of heat, we can use stoichiometry to calculate the enthalpy change for the given amount of CH4:
Enthalpy change (ΔH) = moles of CH4 × ΔH per mole of CH4
= 0.311 mol × (-891 kJ/mol)
= -277.201 kJ
Therefore, the enthalpy change for burning 5.00 g of methane in excess oxygen is -277.201 kJ.
To calculate the enthalpy change for the given reaction, we need to use the equation:
ΔH = (ΣnΔHf(products)) - (ΣnΔHf(reactants))
Where:
ΔH is the enthalpy change
n is the stoichiometric coefficient
ΔHf is the standard enthalpy of formation
First, we need to determine the moles of CH4 (methane) used in the reaction. To do this, we can use the molar mass of CH4.
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.01 g/mol) (H)
Molar mass of CH4 = 16.05 g/mol
Now, we can calculate the moles of CH4:
moles of CH4 = mass of CH4 / molar mass of CH4
moles of CH4 = 5.00 g / 16.05 g/mol
moles of CH4 = 0.311 moles
Next, we can calculate the enthalpy change using the given standard enthalpy of formation values:
ΔH = (ΣnΔHf(products)) - (ΣnΔHf(reactants))
Reactants:
ΔHf(CH4) = 0 kJ/mol (Since it is the elemental form of carbon and hydrogen)
Products:
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
Now, we can calculate the enthalpy change:
ΔH = (2 × ΔHf(CO2)) + (2 × ΔHf(H2O)) - (1 × ΔHf(CH4))
ΔH = (2 × -393.5 kJ/mol) + (2 × -285.8 kJ/mol) - (1 × 0 kJ/mol)
ΔH = -787 kJ/mol
Finally, we can calculate the enthalpy change for the given case where 5.00 g of methane is burned in excess oxygen:
ΔH (for 5.00 g) = ΔH × moles of CH4
ΔH (for 5.00 g) = -787 kJ/mol × 0.311 moles
ΔH (for 5.00 g) = -244.16 kJ
Therefore, the enthalpy change for burning 5.00 g of methane in excess oxygen is approximately -244.16 kJ.
5gCH4/16gCH4= 0.3125moles CH4
(0.3125molesCH4)x(891kJ)=278kJ
Enthalpy change = 278kJ