Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ Express your answer numerically, to four significant figures and in terms of kJ.

ΔH = 130.5 kJ

Explanation:

Hello,

In this case, by using the Hess law, we compute the enthalpy of the required reaction:

C(s) + H2O(g) --> CO(g) + H2(g)

Thus, the first step is to keep the following reaction unchanged:

C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ

Next, we invert and halve this reaction:

2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ

So the enthalpy of reaction is inverted and halved:

CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ

Then, we also invert and halve this reaction:

2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ

So the enthalpy of reaction is inverted and halved as well:

H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ

Finally, we add the three reactions to obtain the required reaction:

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)

= C (s) + H2O → H2 (g) CO (g)

So enthalpy is computed by:

ΔH = -393.5 kJ + 283 kJ + 241.8 kJ

ΔH = 130.5 kJ

Best regards.

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.