Consider the following reaction: 3 O_2(g) + 6 H_2(g) + 6C(s, graphite) rightarro

Question

Consider the following reaction: 3 O_2(g) + 6 H_2(g) + 6C(s, graphite) rightarrow C_6H_12O_6 (s, graphite) What Delta H degree for this reaction? Express your answer to five significant figures. kJ What is Delta S degree for this reaction? Express your answer to five significant figures. When is Delta H_t for a substance equal to zero? Delta H_t is zero for an element in its standard state at 0 K. Delta H_t is zero for an element in its standard state at 25 degree C. Delta H_t is zero for an element in its standard state at 298 15 degree C. Delta H_t is zero for an element in its standard state at 25 K.

Explanation / Answer

Ans part A

delta H for the reaction= delta H of products - delta H of reactants

delta H of products= -1273.3 kJ

delta H of reactants= 3×0 + 6×0 + 6×0 = 0 KJ

delta H = -1273.3 - 0

delta H for the reaction= -1273.3KJ

Part B.

Delta S for the reaction = delta S of products - delta S of reactants

delta S of product = 212.1 KJ

delta S of reactants = (3×205.2) + (6×130.7) +( 6×5.7)

=615.6 +784.2+34.2

=1434 kJ

delta S for the reaction = 212.1 - 1434

Delta S for the reaction = -1221.9 KJ

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