When 4 g of iron is burnt to ferric oxide at a constant pressure, 29.28 kJ of heat is evolved.
The enthalpy of formation of ferric oxide will be-
(At. mass of Fe = 56) ?
1. 81.98 kJ
2. 819.8 kJ
3. 40.99 kJ
4. +819.8 kJ
Hydrolysis of sucrose is given by the following reaction
Sucrose + H2O Glucose + Fructose
If the equilibrium constant (Kc) is 21013 at 300 K, the value of at the same temperature will be:
1. 8.314 J mol–1 K–1300 Kln (21013)
2. 8.314 J mol–1 K–1300 Kln (31013)
3. –8.314 J mol–1 K–1300 Kln (41013)
4. –8.314 J mol–1 K–1300 Kln (21013)
For the reaction, 2Cl(g) Cl2(g), the correct option is:
1.
2.
3.
4.
Assume each reaction is carried out in an open container. For which of the following reactions will
1.
2.
3.
4.
1. \(2R\)
2. \(3R\)
3. \(5R\)
4. \(7R\)
The volume versus temperature graph for two moles of monoatomic gas is shown in the figure. The ratio of work done by the gas to the heat absorbed by it in the process \(A\) to \(B\) is:
1. | \(\dfrac{1}{2}\) | 2. | \(\dfrac{2}{5}\) |
3. | \(\dfrac{3}{7}\) | 4. | \(\dfrac{3}{5}\) |
For the following given equations and values, determine the enthalpy of reaction at 298 K for the reaction:
C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)
H2(g) + F2(g) 2HF(g) = -537 kJ
C(s) + 2F2(g) CF4(g) =-680 kJ
2C(s) + 2H2(g) C2H4(g) = 52 kJ
1. –1165 kJ
2. –2486 kJ
3. +1165 kJ
4. +2486 kJ
Entropy decreases during:
1. Crystallization of sucrose from solution
2. Rusting of iron
3. Melting of ice
4. Vaporization of camphor
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0ºC. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be-
(R = 8.314 J/mol K) (ln 7.5 = 2.01)
1. | q = +208 J, w = -208 J | 2. | q = -208 J, w = -208 J |
3. | q = -208 J, w = + 208 J | 4. | q = +208 J, w = + 208 J |
Given the reaction:
\(2 \mathrm{Cl}(\mathrm{~g}) \rightarrow \mathrm{Cl}_2(\mathrm{~g})\)
What are the values of \(∆\mathrm{H}\) and \(∆\mathrm{S}\), respectively?
1. \(\Delta \mathrm{H}=0, \Delta \mathrm{~S}=-\mathrm{ve}\)
2. \(\Delta \mathrm{H}=0, \Delta \mathrm{~S}=0\)
3. \(\Delta \mathrm{H}=-\mathrm{ve}, \Delta \mathrm{~S}=-\mathrm{ve}\)
4. \(\Delta \mathrm{H}=+\mathrm{ve}, \Delta \mathrm{~S}=+\mathrm{ve}\)