. The standard enthalpy of formation of gas in the above reaction would be-
1. | -92.4 J (mol)-1 | 2. | -46.2 kJ (mol)-1 |
3. | +46.2 J (mol)-1 | 4. | +92.4 kJ (mol)-1 |
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From the following bond energies:
H—H bond energy: 431.37 kJ mol-1
C=C bond energy: 606.10 kJ mol-1
C—C bond energy: 336.49 kJ mol-1
C—H bond energy: 410.50 kJ mol-1
Enthalpy for the reaction,
will be:
1. | 1523.6 kJ mol-1 | 2. | -243.6 kJ mol-1 |
3. | -120.0 kJ mol-1 | 4. | 553.0 kJ mol-1 |
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(298K) of methanol is given by the chemical equation -
1. C(diamond)++2H2CH3OH
2. CH4+CH3OH
3. CO+2H2CH3OH
4. C(graphite)++2H2CH3OH
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The enthalpy of formation of are
–110 kJ , – 393 kJ , 81 kJ and 9.7 kJ respectively.
The value of for the reaction would be-
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Which of the following is not an endothermic reaction?
1. Combustion of methane
2. Decomposition of water
3. Dehydrogenation of ethane or ethylene
4. Conversion of graphite to diamond
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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
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The standard enthalpy of the formation of CH3OH(l) from the following data is:
\(\small{\mathrm{CH}_3 \mathrm{OH}_{(l)}+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}_{(l)} \text {; }}\) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-726 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\) |
\(\small{\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \text {; } }\) \(\Delta_{\mathrm{c}} \mathrm{H}^{\circ}=-393 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\) |
\(\small{\mathrm{H}_{2(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(l)} \text {; } } \) \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\) |
1. | −239 kJ mol−1 | 2. | +239 kJ mol−1 |
3. | −47 kJ mol−1 | 4. | +47 kJ mol−1 |
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