Given the following information determine the reaction Gibbs energy for the formation of silicon dioxide at 298 K

Si(s)+O2(g) →SiO2(s)

ΔfH°(Si(s))=0 kJ mol-1, S°(Si(s))=18.8 J mol-1 K-1

ΔfH°(O2(g))=0 kJ mol-1, S°(O2(g))=205.0 J mol-1 K-1

ΔfH°(SiO2­(s))=-910.7 kJ mol-1, S°(SiO2(s))=41.8 J mol-1 K-1

Answer to TWO decimals places in kJ mol-1

dHo rxn = (n*dHo products) - (n*dHo reactants)

To determine the reaction Gibbs energy (ΔrG°) for the formation of silicon dioxide (SiO2) at 298 K, we can use the equation:

ΔrG° = ΔfG°(SiO2) - ΔfG°(Si) - ΔfG°(O2)

But before we calculate ΔrG°, we need to find the values for ΔfG°(SiO2), ΔfG°(Si), and ΔfG°(O2).

Given information:
ΔfH°(Si(s)) = 0 kJ mol-1
S°(Si(s)) = 18.8 J mol-1 K-1
ΔfH°(O2(g)) = 0 kJ mol-1
S°(O2(g)) = 205.0 J mol-1 K-1
ΔfH°(SiO2­(s)) = -910.7 kJ mol-1
S°(SiO2(s)) = 41.8 J mol-1 K-1

First, let's calculate ΔfG°(Si):
ΔfG°(Si) = ΔfH°(Si) - TΔS°(Si)
= 0 - (298 K * (18.8 J mol-1 K-1 / 1000)) [Converting S from J to kJ]
= 0 - 5.604 kJ mol-1
= -5.604 kJ mol-1

Next, let's calculate ΔfG°(O2):
ΔfG°(O2) = ΔfH°(O2) - TΔS°(O2)
= 0 - (298 K * (205.0 J mol-1 K-1 / 1000)) [Converting S from J to kJ]
= 0 - 61.99 kJ mol-1
= -61.99 kJ mol-1

Finally, let's calculate ΔfG°(SiO2):
ΔfG°(SiO2) = ΔfH°(SiO2) - TΔS°(SiO2)
= -910.7 kJ mol-1 - (298 K * (41.8 J mol-1 K-1 / 1000)) [Converting S from J to kJ]
= -910.7 kJ mol-1 - 12.48 kJ mol-1
= -923.18 kJ mol-1

Now, substituting the values into the equation to calculate ΔrG°:
ΔrG° = ΔfG°(SiO2) - ΔfG°(Si) - ΔfG°(O2)
= (-923.18 kJ mol-1) - (-5.604 kJ mol-1) - (-61.99 kJ mol-1)
= -923.18 kJ mol-1 + 5.604 kJ mol-1 + 61.99 kJ mol-1
= -855.586 kJ mol-1

Therefore, the reaction Gibbs energy for the formation of silicon dioxide at 298 K is approximately -855.59 kJ mol-1.

To determine the reaction Gibbs energy for the formation of silicon dioxide (SiO2) at 298 K, you can use the equation:

ΔG° = ΔH° - TΔS°

where ΔG° is the standard Gibbs energy change, ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change.

Given:
ΔfH°(Si(s)) = 0 kJ mol-1
S°(Si(s)) = 18.8 J mol-1 K-1
ΔfH°(O2(g)) = 0 kJ mol-1
S°(O2(g)) = 205.0 J mol-1 K-1
ΔfH°(SiO2­(s)) = -910.7 kJ mol-1
S°(SiO2(s)) = 41.8 J mol-1 K-1
T = 298 K

First, calculate the standard Gibbs energy change for the formation of silicon dioxide:

ΔG° = ΔH° - TΔS°

ΔG° = [-910.7 kJ mol-1 - (0 kJ mol-1)] - [(298 K) * (41.8 J mol-1 K-1)]

ΔG° = -910.7 kJ mol-1 - 0 kJ mol-1 - 12484 J mol-1

Next, convert the units to kJ mol-1:

ΔG° = -910.7 kJ mol-1 - 0 kJ mol-1 - 12.484 kJ mol-1

ΔG° = -923.184 kJ mol-1

Therefore, the reaction Gibbs energy for the formation of silicon dioxide at 298 K is approximately -923.18 kJ mol-1 (rounded to two decimal places).