Calculate the standard enthalpy change for the reaction
2A+B->2C+D
Use the following data:
Substance
A=-257kj/mol
B=-413kj/mol
C=189kj/mol
D=-475kj/mol
For the reaction given , how much heat is absorbed when 3.00mol of A react?
Is A,B,C,D standard heats of formation?
If so, then by Hess' Law..
Hr=2(C)+D-2(A)-B
That hear of reaction is when 2 moles of A react, so, multiply it by 1.5.
-297KJ/MOL
To calculate the standard enthalpy change for the reaction, we need to determine the change in enthalpy for each of the species involved.
Given:
A = -257 kJ/mol
B = -413 kJ/mol
C = 189 kJ/mol
D = -475 kJ/mol
The enthalpy change for the reaction can be calculated using the formula:
ΔH = (Σn(products) × ΔH(products)) - (Σn(reactants) × ΔH(reactants))
Where Σn is the stoichiometric coefficient and ΔH is the enthalpy change.
For the reaction: 2A + B -> 2C + D
ΔH = (2 × ΔH(C)) + (1 × ΔH(D)) - (2 × ΔH(A)) - (1 × ΔH(B))
= (2 × 189 kJ/mol) + (1 × -475 kJ/mol) - (2 × -257 kJ/mol) - (1 × -413 kJ/mol)
= 378 kJ/mol - 475 kJ/mol + 514 kJ/mol + 413 kJ/mol
= 830 kJ/mol
Therefore, the standard enthalpy change for the reaction is 830 kJ/mol.
To calculate the heat absorbed when 3.00 mol of A react, we can use the formula:
Heat absorbed = ΔH × moles of A
Heat absorbed = 830 kJ/mol × 3.00 mol
= 2490 kJ
Therefore, when 3.00 mol of A react, 2490 kJ of heat is absorbed.
To calculate the standard enthalpy change for a reaction, we need to use the Hess's law and the given enthalpy values of the substances involved in the reaction.
First, we write the balanced equation for the reaction:
2A + B -> 2C + D
According to Hess's law, we can calculate the standard enthalpy change by adding up the enthalpy changes of the individual reactions that produce the desired reaction. In this case, we can consider the following reactions:
1. A -> C
2. B -> D
Given enthalpy values:
A = -257 kJ/mol
B = -413 kJ/mol
C = 189 kJ/mol
D = -475 kJ/mol
Now, let's calculate the standard enthalpy change for each reaction:
1. A -> C:
The coefficient of A in the desired reaction is 2 and the coefficient of C is also 2, so we need to multiply the enthalpy change of this reaction by 2.
Enthalpy change for A -> C = 2 * (-257 kJ/mol) = -514 kJ/mol
2. B -> D:
Enthalpy change for B -> D = -413 kJ/mol
Now, we can sum up the individual enthalpy changes to get the standard enthalpy change for the desired reaction:
Enthalpy change for 2A + B -> 2C + D = -514 kJ/mol + (-413 kJ/mol) = -927 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -927 kJ/mol.
Next, to determine how much heat is absorbed when 3.00 moles of A react, we can use the stoichiometry of the reaction. From the balanced equation, we can see that 2 moles of A react to produce 2 moles of C and 1 mole of B. Therefore, for every 2 moles of A reacted, the heat absorbed will be -927 kJ.
To calculate the heat absorbed when 3.00 moles of A react:
Heat absorbed = (-927 kJ/mol) * (3.00 mol / 2 mol)
Heat absorbed = -1,390.5 kJ
So, when 3.00 moles of A react, approximately 1,390.5 kJ of heat is absorbed.