The value of ΔG° at this temperature is -18034.18 J/mol
Calculation,
Given information
formation constant (Kf)= 1.7 ×
Universal gas constant (R) = 8.314 J/K• mol
Temperature = 25° C = 25 °C + 273 = 300 K
Formula used:
ΔG° = -RT㏑Kf
By putting the valur of R,T, Kf we get the value of ΔG°
ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 ×
ΔG° = -2494.2㏑ 1.7 × = -18034.18 J/mol
So, change in standard Gibbs's free energy is -18034.18 J/mol
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Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice = latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.