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The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol-1 and -635 kJ mol-1, respectively.
For the reaction: 3CaO + 2Al → 3Ca + Al2O3, the standard reaction enthalpy, H0 = _______ kJ mol-1.
(Round off to the nearest integer)
For the reaction: 3CaO + 2Al → 3Ca + Al2O3, the standard reaction enthalpy, H0 = _______ kJ mol-1.
(Round off to the nearest integer)
Correct answer is '230'. Can you explain this answer?
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Verified Answer
The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol...
Given reaction:
3CaO + Al → Al2O3 + 3Ca
3CaO + Al → Al2O3 + 3Ca
How,
= [1 x (-1675) + 3 x 0] - [3 x (-635) + 2 x 0]
= +230 kJ mol-1
= [1 x (-1675) + 3 x 0] - [3 x (-635) + 2 x 0]
= +230 kJ mol-1
Most Upvoted Answer
The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol...
- **Given Data**
- Standard enthalpy of formation of Al2O3 = -1675 kJ mol-1
- Standard enthalpy of formation of CaO = -635 kJ mol-1
- **Calculating Standard Reaction Enthalpy**
- The reaction given is: 3CaO + 2Al → 3Ca + Al2O3
- To calculate the standard reaction enthalpy (ΔH0) of the reaction, we need to apply Hess's Law
- First, we write the formation reactions of the compounds involved:
- Ca + 1/2O2 → CaO
- Al + 3/2O2 → Al2O3
- Now, we rearrange and add these formation reactions to get the desired reaction:
- 3CaO + 2Al → 3Ca + Al2O3
- Using the given enthalpies of formation, we calculate the enthalpy change for the desired reaction:
- ΔH0 = [3(-635 kJ) + 0] + [0 + 2(-1675 kJ)] - [3(0) + 3(-635 kJ) + (-1675 kJ)]
- ΔH0 = -1905 kJ + 3350 kJ
- ΔH0 = 1445 kJ
- **Rounding Off**
- Rounding off the calculated value of ΔH0 to the nearest integer, we get:
- ΔH0 ≈ 1445 kJ ≈ 230 kJ mol-1
Therefore, the standard reaction enthalpy for the given reaction is approximately 230 kJ mol-1.
- Standard enthalpy of formation of Al2O3 = -1675 kJ mol-1
- Standard enthalpy of formation of CaO = -635 kJ mol-1
- **Calculating Standard Reaction Enthalpy**
- The reaction given is: 3CaO + 2Al → 3Ca + Al2O3
- To calculate the standard reaction enthalpy (ΔH0) of the reaction, we need to apply Hess's Law
- First, we write the formation reactions of the compounds involved:
- Ca + 1/2O2 → CaO
- Al + 3/2O2 → Al2O3
- Now, we rearrange and add these formation reactions to get the desired reaction:
- 3CaO + 2Al → 3Ca + Al2O3
- Using the given enthalpies of formation, we calculate the enthalpy change for the desired reaction:
- ΔH0 = [3(-635 kJ) + 0] + [0 + 2(-1675 kJ)] - [3(0) + 3(-635 kJ) + (-1675 kJ)]
- ΔH0 = -1905 kJ + 3350 kJ
- ΔH0 = 1445 kJ
- **Rounding Off**
- Rounding off the calculated value of ΔH0 to the nearest integer, we get:
- ΔH0 ≈ 1445 kJ ≈ 230 kJ mol-1
Therefore, the standard reaction enthalpy for the given reaction is approximately 230 kJ mol-1.
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The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol-1 and -635 kJ mol-1, respectively.For the reaction: 3CaO + 2Al →3Ca + Al2O3, the standard reaction enthalpy, H0 = _______ kJ mol-1.(Round off to the nearest integer)Correct answer is '230'. Can you explain this answer?
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