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Calculate the heat change accompanying the transformation of C(graphite) to C(diamond). You are given:

$straight C left parenthesis graphite right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space space CO subscript 2 left parenthesis straight g right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 393.5 space kJ straight C left parenthesis diamond right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space CO subscript 2 left parenthesis straight g right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 395.4 space kJ$

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212.
With the help of thermochemical equation given below, determine ∆_r.H⁰ at 298K for the following reactions:

$straight C subscript left parenthesis graphite right parenthesis end subscript space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space CH subscript 4 left parenthesis straight g right parenthesis semicolon space space space increment subscript straight r straight H to the power of 0 space equals space ?$
$left parenthesis straight i right parenthesis space straight C subscript graphite space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space CO subscript 2 left parenthesis straight g right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 393.5 space kJ space mol to the power of negative 1 end exponent left parenthesis ii right parenthesis space straight H subscript 2 left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space straight H subscript 2 straight O left parenthesis straight g right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 285.8 space kJ space mol to the power of negative 1 end exponent left parenthesis iii right parenthesis space CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis straight I right parenthesis space space space rightwards arrow space space space CH subscript 4 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis semicolon space space space space increment subscript straight r straight H to the power of 0 space equals space plus 890.3 space kJ thin space mol to the power of negative 1 end exponent$

We aim at the equation

Here we want one mole of C_(graphite) as a reactant, so write equation (i) as such. We want two moles of H₂(g) as a reactant ; so multiply equation (ii) by 2, we want one mole of CH₄(g) as a product, so write equation (iii) as such.

Adding equations (i), (ii) and (iii), we have,

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213. Calculate the enthalpy of combustion of methane, it is given that heat of formation of CO₂(g), H₂O(l) and CH₄(g) are -393 5 kJ mol^–1, –285 .9 kJ mol^–1 and –748 kJ mol^–1respectively.

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214. Calculate the heat of combustion of glucose from the following data:

left parenthesis straight i right parenthesis space straight C left parenthesis graphite right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow CO subscript 2 left parenthesis straight g right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 395.0 space kJ left parenthesis ii right parenthesis space straight H subscript 2 left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space space straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 269.4 space kJ space space space space space space space space space space space space space space space space space space space space space space space space

left parenthesis iii right parenthesis space 6 straight C left parenthesis graphite right parenthesis space plus space 6 straight H subscript 2 left parenthesis straight g right parenthesis space plus space 3 straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space stack straight C subscript 6 straight H subscript 12 straight O subscript 6 left parenthesis straight s right parenthesis semicolon with left parenthesis glucose right parenthesis below

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215. The enthalpy of combustion of methane, graphite and dihydrogen at 298K are –890.3 kJ mol^–1 –393.5 kJ mol^–1 and –285.8 kJ mol^–1 respectively. Enthalpy of formation of CH₄(g) will be
(i) –74.8 kJ mol^–1 (ii) –52 .27 kJ mol^–1(iii) +74.8 kJ mor^–1 (iv) +52 .26 kJ mol^–1

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216. Calculate the enthalpy of formation of acetic acid if its enthalpy of combustion to CO₂(g) and H₂O(l) is –867.0 kJ mol^–1 and the enthalpies of formation of CO₂(g) and H₂O(l) are respectively –393.5 and –285.9 kJ mol^–1.

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217. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:

left parenthesis straight i right parenthesis space CH subscript 3 OH left parenthesis straight l right parenthesis space plus space 3 over 2 straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon space increment subscript straight r straight H to the power of 0 space equals space minus 726 space kJ space mol to the power of negative 1 end exponent

left parenthesis ii right parenthesis space straight C left parenthesis straight s right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space CO subscript 2 left parenthesis straight g right parenthesis semicolon increment straight H to the power of 0 space equals space minus 393 space kJ space mol to the power of negative 1 end exponent space left parenthesis iii right parenthesis space straight H subscript 2 left parenthesis straight g right parenthesis space plus space space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon space increment subscript straight f straight H to the power of 0 space equals space minus 286 space kJ space mol to the power of negative 1 end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

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218.

The combustion of one mole of benzne takes place at 298K and 1 atm. After combustion, CO₂(g) and H₂O (l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ∆_fH° of benzene. Standard enthalpies of formation of CO₂(g) and H₂O(l) are – 393.5 kJ mol^–1 and –285.83 kJ mor^–1respectively.