Given the following information A+B⟶2DΔ
?∘=760.0 kJ
Δ?∘=298.0 J/K
C⟶DΔ
?∘=466.0 kJ
Δ?∘=−188.0 J/K
calculate Δ?∘ at 298 K for the reaction A+B⟶2C
A + B ⟶ 2D
Δ?∘ = Δ?∘ - T Δ?∘
= 760.0 kJ - 298 K * 298.0 J/K
= 671.196 kJ ..................................(1)
C ⟶ D
Δ?∘ = Δ?∘ - T Δ?∘
= 466.0 kJ - 298 K * (−188.0 J/K)
= 522.024 kJ ..................................(2)
Equation (1) - 2*(2).....................
A + B ⟶ 2C
Δ?∘ = 671.196 kJ - 2*522.024 kJ
Δ?∘ = - 372.852 kJ
Answer : Δ?∘ = - 372.8 kJ
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