The reaction of gaseous H2 and liquid Br2 to give gaseous HBr has ΔH = -17.4 kcal/mol (-72.8 kJ/mol) and ΔS = 27.2 (cal/mol⋅K) (114 J/(mol⋅K)). What is the value of ΔG (in kcal and kJ) for the reaction at 297 K ?
dG = dH - TdS
Isn't this just a plug and chug? Don't forget that dS in in cal while dH is in kcal.
Well, let's calculate ΔG, the change in Gibbs free energy, using the formula ΔG = ΔH - TΔS.
Given:
ΔH = -17.4 kcal/mol
ΔS = 27.2 cal/mol⋅K
T = 297 K
First, let's convert ΔH from kcal to cal by multiplying it by 1000:
ΔH = -17.4 kcal/mol * 1000 = -17400 cal/mol
Now, let's convert ΔS from cal/mol⋅K to cal/mol⋅K:
ΔS = 27.2 cal/mol⋅K
Using the given formula:
ΔG = ΔH - TΔS
Substituting the known values:
ΔG = -17400 cal/mol - (297 K * 27.2 cal/mol⋅K)
Simplifying the equation:
ΔG = -17400 cal/mol - 8004 cal/mol
ΔG = -25404 cal/mol
Now, let's convert ΔG from cal/mol to kcal/mol:
ΔG = -25404 cal/mol / 1000 = -25.404 kcal/mol
So, the value of ΔG for the reaction at 297 K is approximately -25.404 kcal/mol (and that's no joke!).
To find the value of ΔG (Gibbs free energy) for the reaction, you can use the equation:
ΔG = ΔH - TΔS
where:
ΔH is the enthalpy change of the reaction,
ΔS is the entropy change of the reaction, and
T is the temperature in Kelvin.
Given:
ΔH = -17.4 kcal/mol
ΔS = 27.2 cal/mol·K
T = 297 K
Converting cal to kcal and J to kJ:
ΔG = -17.4 kcal/mol - (297 K × 27.2 cal/(mol·K) × 1 kcal/1000 cal)
ΔG = -17.4 - 7.9392
ΔG = -25.3392 kcal/mol
Converting kcal to kJ:
ΔG = -25.3392 kcal/mol × 4.184 kJ/kcal
ΔG ≈ -105.89 kJ/mol
Therefore, the value of ΔG for the reaction at 297 K is approximately -25.3392 kcal/mol and -105.89 kJ/mol.
To find the value of ΔG for the reaction at 297 K, we can use the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
Given:
ΔH = -17.4 kcal/mol (-72.8 kJ/mol)
ΔS = 27.2 cal/mol⋅K (114 J/(mol⋅K))
T = 297 K
Let's now calculate ΔG.
Step 1: Convert ΔH from kcal/mol to kJ/mol.
ΔH = -17.4 kcal/mol = -17.4 * 4.184 kJ/mol (since 1 kcal = 4.184 kJ)
ΔH = -72.696 kJ/mol (approximately -72.7 kJ/mol)
Step 2: Substitute the given values into the equation.
ΔG = ΔH - TΔS
= (-72.7 kJ/mol) - (297 K * 114 J/(mol⋅K))
Step 3: Convert the units of ΔS from cal/mol⋅K to kJ/mol⋅K.
ΔS = 27.2 cal/mol⋅K = 27.2 * 0.004184 kJ/mol⋅K (since 1 cal = 0.004184 kJ)
ΔS = 0.113408 kJ/mol⋅K (approximately 0.113 kJ/mol⋅K)
Step 4: Substitute the values into the equation and calculate ΔG.
ΔG = (-72.7 kJ/mol) - (297 K * 0.113 kJ/(mol⋅K))
ΔG = -72.7 - 33.5616
ΔG ≈ -106.2616 kJ/mol
So, the value of ΔG for the reaction at 297 K is approximately -106.3 kJ/mol.
To find ΔG in kcal/mol, we can convert the value by dividing by 4.184 (since 1 kcal = 4.184 kJ):
ΔG ≈ -106.3 kJ/mol / 4.184
ΔG ≈ -25.4 kcal/mol
Therefore, the value of ΔG for the reaction at 297 K is approximately -25.4 kcal/mol.