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References Use the References to access important values if needed for this question. A scientist measures the standard enthalpy change for the following reaction to be 214.2 kJ: CH4(g) + H2O(g 3H2(g) + CO(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(g) is 249 s klinol. Incorrect Feedback Incorrect Correct Answer(s): Incorrect The standard enth-249.9 terms of the standard heats of formation of the products and the eq Arm" 4HPf(products). }1°f(reactants) To determine the Arf for a particular compound scientists often use known heats of reaction and formation. CH+H20(2 3H2(2+CO(G) eq eq req req Moles Species 1rf (kJ/mol) req Products 3 H2(g 0.0 req 1 CO(g)110.5 Reactants1 CH4(g)-74.8 1 H20) Previous Next

Explanation / Answer

Answer:

Given reaction is

CH4(g)+H2O(g)-------->3H2(g)+CO(g)

H°rxn=sum of H° products-sum of H° reactants

H°rxn=[(3×H°H2(g))+(H°CO(g))]-[(H°CH4(g))+(H°H2O(g))]

Substitute given values in the above expression

214.2kj=[(3×0)+(-110.5)]-[(-74.8)+H°H2O(g)]

H°H2O(g)=-214.2-110.5+74.8

H°H2O(g)=-249.9 kJ/mol.

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