Calculate the enthalpy change for the reaction C(s) + 2 H2(g) → CH4(g) ΔH = ? Given the enthalpy values for the following reactions CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = −890.4 kJ C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ H2(g) + 1/2 O2(g) → H2O (g) ΔH = −285.8 kJ (a) −1569.7 kJ (b) +211.1 kJ (c) −1855.5 kJ (d) −74.7 kJ

Video transcript

Calculate the entropy change for the hypothetical reaction given the entropy values for the following reactions. So we need to rearrange the thermo chemical equations, the individual thermo chemical equations to get the final equation. So if we look at reaction one, we see here that hour for our final reaction, we have our entropy change at negative 7 90.2 kilojoules. However, we note that our A B 4 g is on the product side of our final reaction. So you must essentially reverse this entropy. Such that instead of negative 790 we have 0.2 we have positive 7 90.2 kilojoules. Looking at our second reaction, we see that our A which is a solid is on the reactant sign on the for reaction two and for the overall reaction. So we can leave that as is. So we have our positive 7 90.2 kilojoules plus our negative 2 91.5 kilojoules. Lastly, we like our third equation. And we see here that while we have our B two on the reactant side, in both our third reaction and our over reaction, the overall reaction has that raise of the path to the coefficient of two. So we must multiply that NV of reaction three by two such that when we do so he get a value of 351 0.0 kilojoules adding that to our overall equation. Once we add up the individual in the piece, we get an answer of 1 47.7 kilojoules or answer trace B as our final answer overall, I hope this helped. And until next time.