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Q.Given the thermochemical equation, C2H2(g)+52 O2(g)→2CO2(g)+H2O(l)\mathrm{C_2H_2(g)} + \frac{5}{2}\,\mathrm{O_2(g)} \rightarrow 2\mathrm{CO_2(g)} + \mathrm{H_2O}(l), ΔrH0\Delta_r H^0 = -1300 kJ. Write thermochemical equations when i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.

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Step 1. Given: C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l), ΔrH0 = -1300 kJ.

Step 2 (i). Multiplying every coefficient by 2 requires multiplying ΔrH0 by 2 as well (rule v): 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l), ΔrH0 = 2 x (-1300) = -2600 kJ. …

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