Question

# If $$S+O_{2}\longrightarrow SO_{2}$$; $$\Delta H=-398.2\ kJ$$: $$SO_{2}+\dfrac{1}{2}O_{2}\longrightarrow SO_{3}$$; $$\Delta H=-98.7\ kJ$$$$SO_{3}+H_{2}O\longrightarrow H_{2}SO_{4}$$; $$\Delta H=-130.2\ kJ$$ : $$H_{2}+\dfrac{1}{2}O_{2}\longrightarrow H_{2}O$$; $$\Delta H=-227.3\ kJ$$The enthalpy of formation of sulphuric acid at $$298\ K$$ will be?

A
854.4 kJ
B
754.4 kJ
C
650.3 kJ
D
433.7 kJ

Solution

## The correct option is B $$-854.4\ kJ$$Solution:- (A) $$-854.4 \; kJ$$Given:-$$S + {O}_{2} \longrightarrow S{O}_{2} \quad \Delta{H} = -398.2 \; kJ ..... \left( 1 \right)$$$$S{O}_{2} + \cfrac{1}{2} {O}_{2} \longrightarrow S{O}_{3} \quad \Delta{H} = -98.7 \; kJ ..... \left( 2 \right)$$$$S{O}_{3} + {H}_{2}O \longrightarrow {H}_{2}S{O}_{4} \quad \Delta{H} = -130.2 \; kJ ..... \left( 3 \right)$$$${H}_{2} + \cfrac{1}{2} {O}_{2} \longrightarrow {H}_{2}O \quad \Delta{H} = -227.3 \; kJ ..... \left( 4 \right)$$Adding $${eq}^{n} \left( 1 \right), \left( 2 \right), \left( 3 \right) \& \left( 4 \right)$$, we have$${H}_{2} + S + 2 {O}_{2} \longrightarrow {H}_{2}S{O}_{4} \quad \Delta{H} = -854.4 \; kJ$$Hence the enthalpy of formation of sulphuric acid is $$-854.4 \; kJ$$.Chemistry

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