Calculate the δg°rxn using the following information. 2 h2s(g) + 3 o2(g) ⟶ 2 so2(g) + 2 h2o(g) δ g∘ = ? δ hf0 (kj/mol) -20.6 0.00 -296.8 -241.8 s0(j/mol k) 205.8 205.2 248.2 188.8

ΔG° =-990.17 kJ /mol

Explanation:

The equation is:

ΔG° = Δ H° -TΔS°  equation 1

Δ H°rxn = Δ H°f (products)- Δ H°f(reactants)

Δ H°rxn = [[2XΔ H°f(SO₂)]+[2Δ H°f(H₂O)]-[2Δ H°f(H₂S)]+[3Δ H°f(O₂)]

Δ H°rxn = [2X(-296.8)+2(-241.8)]-[2X(-20.6)] = -1036

Δ S°rxn = Δ S°f (products)- Δ S°f(reactants)

Δ S°rxn = [[2XΔ S°f(SO₂)]+[2Δ S°f(H₂O)]-[2Δ S°f(H₂S)]+[3Δ S°f(O₂)]

Δ S°rxn = [2(248.2)+2(188.8)]-[2(205.8)+3(205.2)] = -153.8 J/mol K

      = -0.1538 KJ /mol

Putting values in equation 1  ( T = 298 K)

ΔG° = Δ H° -TΔS° = (-1036)-[(298)(-0.1538)

ΔG° =-990.17 kJ /mol

yes

explanation:

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol