Find the correct equation
E2 - E1 - H2 + H1 = n2RT - n1RT
E2 - E1 - H2 - H1 = n2RT - n1RT
H2 - H1 - E2 + E1 = n2RT - n1RT
H2 - H1 - E2 - E1 = n2RT - n1RT
Assuming enthalpy of combustion of hydrogen at 273 K is -286 kJ and enthalpy of fusion of ice at the same temperature to be +6.0 kJ, calculate enthalpy change during formation of 100 g of ice.
+1622 kJ
- 1622 kJ
+292 kJ
- 292 kJ
For which among the following reactions, change in entropy is less than zero?
Sublimation of iodine
Dissociation of hydrogen
Formation of water
Thermal decomposition of calcium carbonate
What is the amount of work done when two moles of ideal gas is compressed from a volume of 1m3 to 10 dm3 at 300 K against a pressure of 100 kPa?
99 kJ
-99kJ
114.9 kJ
-114.9 kJ
The work done during combustion of 9 × 10-2 kg of ethane, C2H6 (g) at 300 K is (Given R = 8.314 J deg-1 mol-1, atomic mass C = 12, H= 1).
6.236 kJ
-6.236 kJ
18.71 kJ
-18.71 kJ
For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)
The correct set of thermodynamic parameters is
G = 0 and S = -ve
G = 0 and S = +ve
G = +ve and S = 0
G = -ve and S = 0
The standard enthalpy of formation of H2O (l) and Fe2O3 (s) are respectively -286 kJ mol-1 and -824 kJ mol-1. What is the standard enthalpy change for the following reaction?
Fe2O3 (s) + 3H2 (g) → 3H2O (l) + 2Fe (s)
-538 kJ mol-1
+538 kJ mol-1
-34 kJ mol-1
+34 kJ mol-1
C.
-34 kJ mol-1
In the above given reaction,
H (H2O) = -286.00 kJ/ mol
H (Fe2O3) = -824 kJ/ mol
Therefore, HR (Heat of reaction) = ΣH (products) - ΣH (reactants)
HR = -(3 × 286) - (-824)
= -858 + 824
= -34 kJ/ mol