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Author Question: The equilibrium constant for the reaction below is 7.2 10-4 at 298 K and 1 atm. HNO2(aq) + H2O(l) ... (Read 72 times)


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Question 1

The enthalpy change for converting 10.0 g of ice at -25.0 °C to water at 80.0 °C is ________ kJ. The specific heats of ice, water, and steam are and respectively. For H2O, = 6.01 kJ mol-1, and .
◦ 12.28
◦ 6.16
◦ 3870
◦ 7.21
◦ 9.88

Question 2

The equilibrium constant for the reaction below is 7.2 × 10-4 at 298 K and 1 atm.

HNO2(aq) + H2O(l) NO2-(aq) + H3O+(aq)

When [HNO2(aq)] = 1.0 M and [NO2-(aq)] = [H3O+(aq)] = 1.0 × 10-5 M, calculate ΔG.
◦ -39.1 kJ/mol
◦ +17.9 kJ/mol
◦ -17.9 kJ/mol
◦ +39.1 kJ/mol
◦ -23.02 kJ/mol

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Marked as best answer by natalie2426 on Jul 8, 2021


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