Calculate the entropy change in the surroundi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The formation of $1 \ mol$ of $H_2O_{(l)}$ releases $286 \ kJ \ mol^{-1}$ of heat,meaning $q_{sys} = -286 \ kJ \ mol^{-1}$.Since the process is ex

Vedclass · Accepted Answer

$959.73 \ J \ K^{-1} \ mol^{-1}$

Vedclass · Answer

(A) The formation of $1 \ mol$ of $H_2O_{(l)}$ releases $286 \ kJ \ mol^{-1}$ of heat,meaning $q_{sys} = -286 \ kJ \ mol^{-1}$.Since the process is exothermic,the surroundings absorb this heat: $q_{surr} = +286 \ kJ \ mol^{-1} = +286000 \ J \ mol^{-1}$.Standard conditions imply a temperature of $T = 298 \ K$.The entropy change for the surroundings is given by $\Delta S_{surr} = \frac{q_{surr}}{T}$.$\Delta S_{surr} = \frac{286000 \ J \ mol^{-1}}{298 \ K} = 959.73 \ J \ K^{-1} \ mol^{-1}$.

Calculate the entropy change in the surroundings when $1.00 \ mol$ of $H_2O_{(l)}$ is formed under standard conditions. Given: $\Delta_f H^{\theta} = -286 \ kJ \ mol^{-1}$.

Similar Questions

The second law of thermodynamics states that in a cyclic process:

One mole of $NaCl$ $(s)$ on melting absorbed $30.5 \ kJ$ of heat and its entropy increased by $28.8 \ J \ K^{-1}$. The melting point of $NaCl$ is ............. $K$.

The enthalpy change for the transition of liquid water to steam is $37.3 \ kJ \ mol^{-1}$ at $373 \ K$. Find $\Delta S$ in $J \ mol^{-1} \ K^{-1}$.

Which of the following is zero during the adiabatic expansion of a gas?

Statement-$1$: For a process at equilibrium,the total entropy is maximum and the change in total entropy is zero.
Statement-$2$: For a reversible process,if the same amount of heat is supplied to a system at two different temperatures,then the entropy change is less at a lower temperature.

Vedclass Test Series

Exam Paper Generator

Online Exam Module