The standard heats of formation of CCl4(g),H2O(g),CO2(g)andHCl(g)at298Kare−25.5,−57.8,−94.1and−22.1kcalmol−1respectively.ΔH0(298K)forthereactionCCl4(g)+2H2O(g)→CO2(g)+4HCl(g)
A
+82.8kcalmol−1
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B
−82.8kcalmol−1
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C
+41.4kcalmol−1
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D
−41.4kcalmol−1
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Solution
The correct option is D−41.4kcalmol−1 We are given the standard heat of formation of reactant and products.
we know that, ΔH0r=∑ΔfH0(products)−∑ΔfH0(reactants)
putting all the values, =−94.1+4(−22.1)−(−25.5−2×57.8)kcalmol−1=−41.4kcalmol−1