Using the data in the table below, calculate the standard enthalpy change, in kJ mol-', for the reaction between carbon
disulfide,CS2, and oxygen shown in the following equation.
CS,(g) + 302(g) → CO₂(g) + 250₂(g)
Substance
CS₂(g)
CO₂(g)
SO₂(g)
Standard enthalpy change of formation, AH,/kJ mol-¹
+110
-390
-290
A) -570
B )-790
C )-860
D )-1080
To calculate the standard enthalpy change for the reaction, we can use the equation:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
Where:
ΔH is the standard enthalpy change for the reaction
ΣnΔHf(products) is the sum of the standard enthalpy changes of formation for the products, multiplied by their stoichiometric coefficients
ΣnΔHf(reactants) is the sum of the standard enthalpy changes of formation for the reactants, multiplied by their stoichiometric coefficients
In this case, the stoichiometric coefficients for CS2, O2, CO2, and SO2 are 1, 3, 1, and 1 respectively.
Plugging in the values from the table:
ΔH = [1*(-390) + 1*(0)] - [1*(110) + 3*(0)]
ΔH = -390 - 110 = -500 kJ/mol
Therefore, the correct answer is not listed.
To calculate the standard enthalpy change for the reaction, we need to consider the enthalpy changes of formation for each substance involved in the reaction:
Reactants:
CS₂(g) = +110 kJ/mol
O₂(g) = 0 kJ/mol (since it is in its standard state)
Products:
CO₂(g) = -390 kJ/mol
SO₂(g) = -290 kJ/mol
Next, we need to determine the stoichiometric coefficients for each substance in the balanced equation:
CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g)
By comparing the coefficients of the given balanced equation with the standard enthalpy changes, we see that 1 mole of CS₂ requires 3 moles of O₂ to react, producing 1 mole of CO₂ and 2 moles of SO₂.
Since we need 3 moles of O₂ for the reaction, the enthalpy change for this part of the reaction will be 3 * 0 = 0 kJ/mol.
The enthalpy change for the formation of CO₂ will be -390 kJ/mol, and for the formation of SO₂, it will be -290 kJ/mol.
Adding up all these enthalpy changes:
ΔH = (0 kJ/mol) + (-390 kJ/mol - (+110 kJ/mol)) + 2(-290 kJ/mol) = -790 kJ/mol
Therefore, the standard enthalpy change for the reaction between carbon disulfide, CS₂, and oxygen is -790 kJ/mol.
The correct answer is B) -790.
To calculate the standard enthalpy change for the reaction, we can use the following principle:
Standard Enthalpy Change = ∑(moles of products * standard enthalpy change of formation) - ∑(moles of reactants * standard enthalpy change of formation)
First, we need to determine the number of moles for each compound. Looking at the balanced equation:
CS2(g) + 3O2(g) → CO2(g) + 2SO2(g)
We can see that there is 1 mole of CS2, 3 moles of O2, 1 mole of CO2, and 2 moles of SO2 involved in the reaction.
Next, we can use the standard enthalpy change of formation values provided in the table to calculate the enthalpy change:
Enthalpy Change = (1 mol * (-390 kJ/mol)) + (1 mol * 3 mol * (-290 kJ/mol)) - (1 mol * (+110 kJ/mol))
Simplifying the equation:
Enthalpy Change = -390 kJ/mol - 870 kJ/mol - 110 kJ/mol
Enthalpy Change = -1370 kJ/mol
Therefore, the standard enthalpy change for the reaction between carbon disulfide, CS2, and oxygen is -1370 kJ/mol.
None of the given options (-570, -790, -860, -1080) match this calculated value, so none of the provided options are correct.