-->

Thermodynamics

Question
CBSEENCH11006209

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

Solution

We aim at the equation
 straight C subscript left parenthesis graphite right parenthesis end subscript space plus space 2 straight H subscript 2 space space space rightwards arrow space 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 circled dash space equals space ?
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.
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 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 2 straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space space 2 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 2 left parenthesis negative 285.8 space straight K space kJ space mol to the power of negative 1 end exponent right parenthesis
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 equals negative 571.6 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 l 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 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 plus 890.3 space kJ space mol to the power of negative 1 end exponent space space
Adding equations (i), (ii) and (iii), we have,
    space straight C subscript graphite space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space space CH subscript 4 left parenthesis straight g right parenthesis
increment subscript straight r straight H to the power of 0 space equals space minus 393.5 space minus 571.6 space plus space 890.3
space space space space space space space space space equals space minus 74.8 space kJ space mol to the power of negative 1 end exponent