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Consider the following numbered processes:rni. a → 2b ∆h1rnrnii. b → c + d ∆h2rnrniii. e → 2d ∆h3rnrn∆h for the process a → 2c + e is which of the following?
Accepted Answer
To find the enthalpy change (∆H) for the process A → 2C + E, we need to manipulate the given processes to match the desired reaction. Here's how:
1. Reverse process II: C + D → B with ∆H = -∆H2
2. Multiply process II by 2: 2C + 2D → 2B with ∆H = -2∆H2
3. Add process I and the modified process II: A + 2C + 2D → 2B + 2B with ∆H = ∆H1 - 2∆H2
4. Simplify: A + 2C + 2D → 4B with ∆H = ∆H1 - 2∆H2
5. Reverse process III: 2D → E with ∆H = -∆H3
6. Add the result from step 4 and process III: A + 2C + 2D + 2D → 4B + E with ∆H = ∆H1 - 2∆H2 - ∆H3
7. Simplify: A + 2C → 4B + E with ∆H = ∆H1 - 2∆H2 - ∆H3
8. Since we want A → 2C + E, reverse the final equation: 4B + E → A + 2C with ∆H = -(∆H1 - 2∆H2 - ∆H3)
9. Simplify: A → 2C + E with ∆H = -∆H1 + 2∆H2 + ∆H3
Therefore, the enthalpy change for the process A → 2C + E is -∆H1 + 2∆H2 + ∆H3.
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